C理论A_编程练习1-顺序结构

主要是一些基础语法，输入输出，和对类型的理解。

7-1 计算整数各位数字之和

这里直接用循环实现了。

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#include<stdio.h>
int main() {
    int a, b = 0;
    scanf("%d", &a);
    while (a)
        b += a % 10, a /= 10;
    printf("%d", b);
    return 0;
}

7-2 计算代数表达式

使用内置的数学函数解决

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#include<stdio.h>
#include<math.h>
#define PI 3.14159
#define to(x) (x * PI / 180)
int main() {
    double x;
    scanf("%lf", &x);
    printf("%.2lf", sqrt((sin(to(60)) + 1) * (sin(to(30)) + 1) / cos(to(x))));
    return 0;
}

7-3 A乘以B

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#include<stdio.h>
int main() {
    int a, b;
    scanf("%d%d", &a, &b);
    printf("%d", a * b);
    return 0;
}

7-4 是不是太胖了

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#include<stdio.h>
int main() {
    int h;
    scanf("%d", &h);
    printf("%.1f", (h - 100) * 0.9f * 2);
    return 0;
}

7-5 整数四则运算

不会函数和指针者慎重参考。

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#include<stdio.h>
char op[] = {'+', '-', '*', '/'};
typedef int (*expr)(int, int);
int plus(int a, int b) {return a + b;}
int minus(int a, int b) {return a - b;}
int mult(int a, int b) {return a * b;}
int div(int a, int b) {return a / b;}
expr funcs[] = {plus, minus, mult, div};
int main() {
    int a, b;
    scanf("%d%d", &a, &b);
    for (int i = 0;i < 4;i++) {
        printf("%d %c %d = %d\n", a, op[i], b, funcs[i](a, b));
    }
    return 0;
}

7-6 杨辉三角

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#include<stdio.h>
#define N 5
int a[2][N + 10];
int main() {
    for (int i = 1;i <= N;i++) {
        for (int j = 1;j <= i;j++) {
            if (j == 1 || j == i) {
                a[i % 2][j] = 1;
            } else {
                a[i % 2][j] = a[(i-1) % 2][j - 1] + a[(i-1) % 2][j];
            }
            printf("%3d", a[i % 2][j]);
        }
        putchar('\n');
    }
    return 0;
}

7-7 计算物体自由下落的距离

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#include<stdio.h>
int main() {
    int t;
    scanf("%d", &t);
    printf("height = %.2f", 0.5f * 9.8f * t * t);
    return 0;
}

7-8 厘米换算英尺英寸

主要考察算数运算和类型强制转换

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#include<stdio.h>
int main() {
    int cm;
    scanf("%d", &cm);
    float tot = cm / 100.0f / 0.3048f;
    printf("%d %d", (int)tot, (int)((tot - (int)tot) * 12));
    return 0;
}

C理论A_编程练习2-选择结构

7-1 A除以B

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#include<stdio.h>
int main() {
    int a, b;
    scanf("%d%d", &a, &b);
    if (b > 0) {
        printf("%d/%d=%.2f", a, b, a*1.0f/b);
    } else if (b < 0) {
        printf("%d/(%d)=%.2f", a, b, a*1.0f/b);
    } else {
        printf("%d/%d=Error", a, b);
    }
    return 0;
}

7-2 出租车计价

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#include<stdio.h>
int main() {
    float d, fee;
    int t;
    scanf("%f%d", &d, &t);
    if (d <= 3) {
        fee = 10;
    } else if (d <= 10) {
        fee = 10 + (d - 3) * 2;
    } else {
        fee = 10 + (d - 3) * 2 + (d - 10) * 1;
    }
    fee += t / 5 * 2;
    printf("%.0f", fee);
    return 0;
}

7-3 输出英文字母的前驱和后继字母

考虑ascii码

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#include<stdio.h>
int main() {
    char s[10];
    scanf("%s", s);
    char c = s[0];
    char prv = c - 1, nxt = c + 1;
    if (c >= 'a' && c <= 'z') {
        if (prv < 'a') {
            prv = 'z';
        }
        if (nxt > 'z') {
            nxt = 'a';
        }
    } else {
        if (prv < 'A') {
            prv = 'Z';
        }
        if (nxt > 'Z') {
            nxt = 'A';
        }
    }
    printf("%c %d\n%c %d", prv, (int)prv, nxt, (int)nxt);
    return 0;
}

7-4 后天

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#include<stdio.h>
int main() {
    int d;
    scanf("%d", &d);
    printf("%d", (d + 2) % 8 + (d + 2) / 8);
    return 0;
}

7-5 后天日期

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#include<stdio.h>
int t[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int getDay(int y, int m) {
    if (m == 2 && ((y % 4 == 0 && y % 100 != 0) || (y % 400 == 0))) {
        return 29;
    }
    return t[m];
}
int main() {
    int y, m, d;
    scanf("%d-%d-%d", &y, &m, &d);
    d += 2;
    int tot = getDay(y ,m);
    if (d > tot) {
        d -= tot;
        m ++;
    }
    if (m > 12) {
        m -= 12;
        y ++;
    }
    printf("%d.%02d.%02d", y, m, d);
    return 0;
}

7-6 计算工资

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#include<stdio.h>
#define max(a, b) (a > b ? a : b)
int main() {
    int y, h;
    scanf("%d%d", &y, &h);
    int b = y >= 5 ? 50 : 30;
    printf("%.2f", (h * b + b * max(0, h - 40) * 0.5f));
    return 0;
}

7-7 能买手机吗？

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#include<stdio.h>
#include<math.h>
int main() {
    int p;
    scanf("%d", &p);
    if (p > 40 * 20 * 4) {
        float l = (p - 40 * 20 * 4) / 40.0f;
        if (l > 4 * 10) {
            printf("需加班%.0f小时,买不起", ceil(l));
        } else {
            printf("需加班%.0f小时,可购买", ceil(l));
        }
    } else {
        puts("不需加班,可购买");
    }
    return 0;
}

7-8 谁是赢家

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#include<stdio.h>
int main() {
    int pa, pb, c, b = 0;
    scanf("%d%d", &pa, &pb);
    for (int i = 0;i < 3;i++) {
        scanf("%d", &c);
        b += c;
    }
    if ((pb > pa && b > 0) || (pb < pa && b == 3)) {
        printf("The winner is b: %d + %d", pb, b);
    } else {
        printf("The winner is a: %d + %d", pa, 3 - b);
    }
    return 0;
}

7-9 比较大小

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#include<stdio.h>
int main() {
    int a[10];
    for (int i = 0;i < 3;i++) {
        scanf("%d", &a[i]);
    }
    for (int i = 0;i < 3;i++) {
        for (int j = 0;j < 2 - i;j++) {
            if (a[j] > a[j + 1]) {
                a[j] ^= a[j + 1];
                a[j + 1] ^= a[j];
                a[j] ^= a[j + 1];
            }
        }
    }
    for (int i = 0;i < 3;i++) {
        printf("->%d" + (i == 0) * 2, a[i]);
    }
    return 0;
}

C理论A_编程练习3-循环结构

7-1 求平方根序列前N项和

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#include<stdio.h>
#include<math.h>
int main() {
    int n;
    scanf("%d", &n);
    double s = 0;
    for (int i = 1;i <= n;i++) {
        s += sqrt(i);
    }
    printf("sum = %.2lf", s);
    return 0;
}

7-2 统计整数的位数

为什么写成do-while？因为0的特判。

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#include<stdio.h>
int main() {
    int n, l = 0;
    scanf("%d", &n);
    do {
        n /= 10;
        l ++;
    } while (n);
    printf("It contains %d digits.", l);
    return 0;
}

7-3 输出闰年

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#include<stdio.h>
int main() {
    int n;
    scanf("%d", &n);
    if (n <= 2000 || n > 2100) {
        puts("Invalid year!");
    } else {
        int f = 1;
        for (int i = 2001;i <= n;i++) {
            if ((i % 4 == 0 && i % 100 != 0) || (i % 400 == 0)) {
                printf("%d\n", i);
                f = 0;
            }
        }
        if (f) {
            puts("None");
        }
    }
    return 0;
}

7-4 求分数序列前N项和

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#include<stdio.h>
int main() {
    int n;
    scanf("%d", &n);
    double s = 0;
    double x = 2, y = 1;
    while (n--) {
        s += x / y;
        double t = x;
        x = x + y;
        y = t;
    }
    printf("%.2lf", s);
    return 0;
}

7-5 求给定精度的简单交错序列部分和

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#include<stdio.h>
#define ll long long
int main() {
    double eps;
    scanf("%lf", &eps);
    double s = 0, c;
    ll i = 0;
    do {
        c = 1.0 / (1 + 3 * i);
        s += (i&1) ? -c : c;
        i++;
    } while (c > eps);
    printf("sum = %.6lf", s);
    return 0;
}

7-6 求cosx的近似值

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#include<stdio.h>
int main() {
    double x;
    int n;
    scanf("%lf%d", &x, &n);
    double s = 0, m = 1, y = 1;
    for (int i = 0;i <= n;i++) {
        s += y / m * ((i&1) ? -1 : 1);
        y *= x * x;
        m *= (2 * i + 1) * (2 * i + 2);
    }
    printf("cos(%.6lf)=%.6lf", x, s);
    return 0;
}

7-7 统计学生成绩

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#include<stdio.h>
int l[] = {90, 80, 70, 60, 0};
int cs[5];
int main() {
    int n, c;
    scanf("%d", &n);
    for (int i = 0;i < n;i++) {
        scanf("%d", &c);
        for (int j = 0;j < 5;j++) {
            if (c >= l[j]) {
                cs[j]++;
                break;
            }
        }
    }
    for (int i = 0;i < 5;i++) {
        printf(" %d" + !i, cs[i]);
    }
    return 0;
}

7-8 最佳情侣身高差

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#include<stdio.h>
int main() {
    int n;
    scanf("%d", &n);
    double c;
    char s[10];
    while (n--) {
        scanf("%s%lf", s, &c);
        printf("%.2lf\n", s[0] == 'M' ? c / 1.09 : c * 1.09);
    }
    return 0;
}

7-9 简写转全称

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#include<stdio.h>
char s[1024];
int main() {
    scanf("%s", s);
    int i = 0;
    while (s[i] != '\0') {
        switch(s[i++]) {
            case 'c':case 'C':
                puts("BEIJING OLYMPIC GAMES");
                break;
            case 'j':case 'J':
                puts("JAPAN WORLD CUP");
                break;
            case 'k':case 'K':
                puts("KOREA WORLD CUP");
                break;
            default:
                putchar(s[i - 1]);
                putchar('\n');
        }
    }
    return 0;
}

7-10 约分最简分式

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#include<stdio.h>
int gcd(int a, int b) {
    if (b) {
        return gcd(b, a % b);
    }
    return a;
}
int main() {
    int a, b;
    scanf("%d/%d", &a , &b);
    int x = gcd(a, b);
    printf("%d/%d", a / x, b / x);
    return 0;
}

7-11 猴子吃桃问题

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#include<stdio.h>
int main() {
    int n, t = 1;
    scanf("%d", &n);
    while (--n) {
        t = (t + 1) * 2;
    }
    printf("%d", t);
    return 0;
}

7-12 特殊a串数列求和

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#include<stdio.h>
int main() {
    int a, n;
    scanf("%d%d", &a, &n);
    int s = 0, d = a;
    while (n--) {
        s += d;
        d = d * 10 + a;
    }
    printf("s = %d", s);
    return 0;
}

7-13 穷举法搬运砖块问题

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#include<stdio.h>
int main() {
    int n, f = 1;
    scanf("%d", &n);
    for (int i = 0;i <= n / 3;i++) {
        int r = n - 3 * i;
        for (int j = 0;j <= r / 2 && i + j <= n;j++) {
            int rr = r - 2 * j;
            if (i + j + rr*2 == n) {
                printf("men=%d women=%d child=%d\n", i, j, rr*2);
                f = 0;
            }
        }
    }
    if (f) {
        puts("No solution!");
    }
    return 0;
}

7-14 数字金字塔

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#include<stdio.h>
int main() {
    int t, n;
    scanf("%d", &t);
    while (t--) {
        scanf("%d", &n);
        for (int i = 1;i <= n;i++) {
            for (int j = 0;j < (n - i) * 2;j++) {
                putchar(' ');
            }
            for (int j = 0;j < 1 + (i - 1) * 2;j++) {
                printf("%d ", i);
            }
            putchar('\n');
        }
    }
    return 0;
}

7-15 输出N以内的所有素数

使用素数筛法

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#include<stdio.h>
short p[600];
int main() {
    int t;
    scanf("%d", &t);
    p[1] = 1;
    for (int i = 2;i <= t;i++) {
        if (!p[i]) {
            for (int j = i * 2;j <= t;j += i) {
                p[j] = 1;
            }
        }
    }
    int cnt = 0;
    for (int i = 1;i <= t;i++) {
        if (!p[i]) {
            printf("%5d", i);
            cnt ++;
            if (cnt % 8 == 0) {
                putchar('\n');
            }
        }
    }
    return 0;
}

C理论A_编程练习4-数组

7-1 冒泡法排序

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#include<stdio.h>
int main() {
    int a[110], n, k;
    scanf("%d%d", &n, &k);
    for (int i = 0;i < n;i++) {
        scanf("%d", &a[i]);
    }
    for (int i = 0;i < k;i++) {
        for (int j = 0;j < n - 1 - i;j++) {
            if (a[j] > a[j + 1]) {
                a[j] ^= a[j + 1];
                a[j + 1] ^= a[j];
                a[j] ^= a[j + 1];
            }
        }
    }
    for (int i = 0;i < n;i++) {
        printf(" %d" + !i, a[i]);
    }
    return 0;
}

7-2 评委打分

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#include<stdio.h>
int main() {
    int n, m[11] = {};
    for (int i = 1;i <= 10;i++) {
        scanf("%d", &m[i]);
    }
    scanf("%d", &n);
    while (n--) {
        scanf("%d", &m[0]);
        m[m[0]] += 10;
    }
    for (int i = 1;i <= 10;i++) {
        printf(" %d" + (i == 1), m[i]);
    }
    return 0;
}

7-3 组合数的和

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#include<stdio.h>
int main() {
    int n, s = 0, c;
    scanf("%d", &n);
    for (int i = 0;i < n;i++) {
        scanf("%d", &c);
        s += c * 11 * (n - 1);
    }
    printf("%d", s);
    return 0;
}

7-4 找不同

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#include<stdio.h>
struct {
    int cnt;
    int time;
} d[100100];
int main() {
    int n, c, max = 0;
    scanf("%d", &n);
    for (int i = 1;i <= n;i++) {
        scanf("%d", &c);
        if (c > max) {
            max = c;
        }
        d[c].cnt ++;
        if (d[c].cnt == 1) {
            d[c].time = i;
        }
    }
    int m = -1, ts = 11451419;
    for (int i = 1;i <= max;i++) {
        if (d[i].cnt == 1) {
            if (m == -1 || d[i].time < ts) {
                m = i;
                ts = d[i].time;
            }
        }
    }
    if (m == -1) {
        puts("None");
    } else {
        printf("%d", m);
    }
    return 0;
}

7-5 利用二分查找搜寻所有待查找数据

有点申必。次数对不上，后来改成这b样踩过

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#include<stdio.h>
int a[50], n, cnt;
int binarySearch(int x) {
    cnt = 0;
    int l = 0, r = n, lastmid = -1;
    while (l < r) {
        cnt ++;
        int mid = (l + r) / 2;
        if (mid == lastmid) {
            return -1;
        }
        if (a[mid] > x) {
            r = mid;
        } else if (a[mid] < x) {
            l = mid;
        } else {
            return mid;
        }
        lastmid = mid;
    }
    if (a[l] == x) {
        return l;
    }
    return -1;
}
int main() {
    int x;
    scanf("%d", &n);
    for (int i = 0;i < n;i++) {
        scanf("%d", &a[i]);
    }
    scanf("%d", &x);
    int left = binarySearch(x);
    if (left == -1) {
        puts("not found");
    } else {
        printf("查找次数%d\n", cnt);
        int right = left + 1;
        while (left > 0 && a[left - 1] == x) {
            left --;
        }
        while (right < n && a[right] == x) {
            right ++;
        }
        for (int i = left;i < right;i++) {
            printf("位置:%d\n", i);
        }
    }
    return 0;
}

7-6 判断上三角矩阵

根本用不上数组，笑死

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#include<stdio.h>
int main() {
    int t, n, c;
    scanf("%d", &t);
    while (t--) {
        scanf("%d", &n);
        int flag = 1;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                scanf("%d", &c);
                if (j < i) {
                    flag = flag && !c;
                }
            }
        }
        puts(flag ? "YES" : "NO");
    }
    return 0;
}

7-7 方阵循环右移

算算公式就好，不用真的右移

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#include<stdio.h>
int main() {
    int m, n, a[20];
    scanf("%d%d", &m, &n);
    for (int i = 0;i < n;i++) {
        for (int j = 0;j < n;j++) {
            scanf("%d", &a[j]);
        }
        for (int j = 0;j < n;j++) {
            printf("%d ", a[(n * (m / n + 1) + j - m) % n]);
        }
        putchar('\n');
    }
    return 0;
}

7-8 TicTacToe游戏–胜负判定

代码重用

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#include<stdio.h>
char m[10][10];
int check(char x) {
    if (m[0][0] == x && m[1][1] == x && m[2][2] == x) {
        return 1;
    }
    if (m[0][2] == x && m[1][1] == x && m[2][0] == x) {
        return 1;
    }
    for (int i = 0;i < 3;i++) {
        int x1 = 1, x2 = 1;
        for (int j = 0;j < 3;j++) {
            x1 = x1 && m[i][j] == x;
            x2 = x2 && m[j][i] == x;
        }
        if (x1 | x2) {
            return 1;
        }
    }
    return 0;
}
int main() {
    for (int i = 0;i < 3;i++) {
        scanf("%s", m[i]);
    }
    if (check('O')) {
        puts("O win!");
    } else if (check('X')) {
        puts("X win!");
    } else {
        puts("No one win!");
    }
    return 0;
}

7-9 2048游戏模拟（2）–向下移位合并

错了几次确实是我忘了2048怎么玩的。。。

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#include<stdio.h>
int m[5][5];
int x[5][5];
int main() {
    for (int i = 0;i < 16;i++) {
        scanf("%d", &m[i / 4][i % 4]);
    }
    for (int i = 0;i < 4;i++) {
        for (int j = 2;j >= 0;j--) {
            if (!m[j][i]) {
                continue;
            }
            int k = j;
            while (k < 3 && !m[k + 1][i]) k++;
            if (k < 3 && m[k + 1][i] == m[j][i] && !x[k + 1][i]) {
                m[k + 1][i] *= 2;
                m[j][i] = 0;
                x[k + 1][i] = 1;
            } else if (k != j){
                m[k][i] = m[j][i];
                m[j][i] = 0;
            }
        }
    }
    for (int i = 0;i < 4;i++) {
        for (int j = 0;j < 4;j++) {
            printf(" %d" + !j, m[i][j]);
        }
        putchar('\n');
    }
    return 0;
}

7-10 二进制数据转换成十进制数

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#include<stdio.h>
#include<string.h>
int main() {
    char str[32];
    int x = 0, d = 1;
    scanf("%s", str);
    int len = strlen(str);
    for (int i = len - 1;i >= 0;i--) {
        x += (str[i] - '0') * d;
        d *= 2;
    }
    printf("%d", x);
    return 0;
}

7-11 英文单词排序

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#include<stdio.h>
#include<string.h>
struct S {
    char text[20];
    int len;
} s[25];
int pos[25];
int cmp(int i, int j) {
    int r = s[pos[i]].len - s[pos[j]].len;
    if (r == 0) {
        r = pos[i] - pos[j];
    }
    return r;
}
int main() {
    int n = 0;
    while (1) {
        scanf("%s", s[n].text);
        if (!strcmp(s[n].text, "#")) {
            break;
        }
        s[n].len = strlen(s[n].text);
        pos[n] = n;
        n ++;
    }
    for (int i = 0;i < n;i++) {
        for (int j = 0;j < n - i - 1;j++) {
            if (cmp(j, j + 1) > 0) {
                pos[j] ^= pos[j + 1];
                pos[j + 1] ^= pos[j];
                pos[j] ^= pos[j + 1];
            }
        }
    }
    for (int i = 0;i < n;i++) {
        printf("%s ", s[pos[i]].text);
    }
    return 0;
}

7-12 统计单词的长度

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#include<stdio.h>
#include<string.h>
char buf[1024];
int main() {
    int f = 1;
    while (scanf("%s", buf) != -1) {
        printf("%d ", strlen(buf));
        f = 0;
    }
    if (f) {
        printf("0 ");
    }
    return 0;
}

7-13 来验证我们的身份证吧

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#include<stdio.h>
const int weights[] = { 7, 9, 10, 5, 8, 4, 2, 1, 6, 3, 7, 9, 10, 5, 8, 4, 2 };
const char checksum[] = "1 0 X 9 8 7 6 5 4 3 2 ";
int main() {
    int n, f = 1;
    char str[20];
    scanf("%d", &n);
    while (n--) {
        scanf("%s", str);
        int sum = 0;
        for (int i = 0;i < 17;i++) {
            sum += (str[i] - '0') * weights[i];
            sum %= 11;
        }
        if (checksum[sum * 2] != str[17]) {
            puts(str);
            f = 0;
        }
    }
    if (f) {
        puts("全部正确!");
    }
    return 0;
}

7-14 将整数按三位分节

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#include<stdio.h>
#include<string.h>
int main() {
    char str[32];
    scanf("%s", str);
    int len = strlen(str);
    int skip = len % 3;
    for (int i = 0;i < len;i++) {
        if (i >= skip && i != 0 && (i - skip) % 3 == 0) {
            putchar(',');
        }
        putchar(str[i]);
    }
    return 0;
}

7-15 求完全对称日

顺便喷一句，一开始测试数据完全没有考虑闰年。我的内心是操蛋的。

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#include<stdio.h>
const int ds[] = {
    0, 31, 28, 31, 30, 31, 30, 31,
    31, 30, 31, 30, 31
};
int main() {
    int m, n;
    scanf("%d%d", &m, &n);
    int flag = 1;
    for (int i = m;i <= n;i++) {
        int y = i;
        int m = y % 100;
        m = (m % 10) * 10 + m / 10;
        int d = y / 100;
        d = (d % 10) * 10 + d / 10;
        if (m >= 1 && m <= 12 && d > 0) {
            int d2 = (m == 2 && ((y % 4 == 0 && y % 100 != 0) || y % 400 == 0)) ? 29 : ds[m];
            if (d <= d2) {
                printf("%04d.%02d.%02d\n", y, m, d);
                flag = 0;
            }
        }
    }
    if (flag) {
        puts("None");
    }
    return 0;
}

C理论A_编程练习5-函数与指针

6-1 求实数和的函数

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float sum(float data[], int n) {
    float s = 0;
    for (int i = 0;i < n;i++) {
        s += data[i];
    }
    return s;
}

6-2 求解一元二次方程实根的函数

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int rootOfEquation(double a, double b, double c, double *x1, double *x2) {
    double delta = b * b - 4 * a * c;
    if (delta < 0) {
        return 0;
    } else {
        double s1 = (-b + sqrt(delta)) / (2 * a);
        double s2 = (-b - sqrt(delta)) / (2 * a);
        *x1 = s1 > s2 ? s1 : s2;
        *x2 = s1 > s2 ? s2 : s1;
        return delta == 0 ? 1 : 2;
    }
}

6-3 求集合数据的均方差

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double Avg ( int N, int  data[] ) {
    double s = 0;
    for (int i = 0;i < N;i++) {
        s += data[i];
    }
    return s / N;
}
double StdDev( int N, int  data[] ) {
    double avg = Avg(N, data);
    double s = 0;
    for (int i = 0;i < N;i++) {
        s += (data[i] - avg) * (data[i] - avg);
    }
    return sqrt(s / N);
}

6-4 计算Fibonacci数列每一项时所需的递归调用次数

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int count;

long Fib(int a) {
    count ++;
    if (a == 1 || a == 2) {
        return 1;
    }
    return Fib(a - 1) + Fib(a - 2);
}

6-5 字符串加密

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void cryptograp(char ch[],int n) {
    for (int i = 0;i < n;i++) {
        char salted = ch[i] + 5;
        if (ch[i] >= 'a' && ch[i] <= 'z' && salted > 'z') {
            salted = 'a' + salted - 'z' - 1;
        } else if (ch[i] >= 'A' && ch[i] <= 'Z' && salted > 'Z') {
            salted = 'A' + salted - 'Z' - 1;
        } else if(!((ch[i] >= 'A' && ch[i] <= 'Z') || (ch[i] >= 'a' && ch[i] <= 'z'))){
            continue;
        }
        ch[i] = salted;
    }
}

6-6 万年历显示函数

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void ShowDate(int y, int m) {
    puts("***********************************");
    puts("  Sun  Mon  Tue  Wen  The  Fri  Sta");
    int days = GetDaysofMonth(y, m);
    int loc = GetFirstDayInTable(y, m);
    for (int i = 1;i <= days;i++) {
        if (i == 1) {
            for (int j = 0;j < loc;j++) {
                printf("     ");
            }
        }
        printf("%5d", i);
        if ((loc + i) % 7 == 0) {
            putchar('\n');
        }
    }
    if ((loc + days) % 7 != 0) {
        putchar('\n');
    }
    puts("***********************************");
}

C理论A_编程练习6-结构体与共用体

6-1 为通讯录排序

注：本题为函数题。

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struct Records {
    char name[16], birth[10], tele[25];
};
#include<string.h>
int cmp(struct Records *a, struct Records *b) {
    return strcmp(a->name, b->name);
}
char buf[50];
void strswp(char *a, char *b) {
    strcpy(buf, a);
    strcpy(a, b);
    strcpy(b, buf);
}
void sswap(struct Records *a, struct Records *b) {
    strswp(a->name, b->name);
    strswp(a->birth, b->birth);
    strswp(a->tele, b->tele);
}
void my_sort(struct Records *addr, int N) {
    for (int i = 0;i < N;i++) {
        for (int j = 0;j < N - i - 1;j++) {
            if (cmp(addr + j, addr + j + 1) > 0) {
                sswap(addr + j + 1, addr + j);
            }
        }
    }
}

7-1 查找书籍

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#include <stdio.h>
struct Book {
    double price;
    char name[50];
} books[50];
int main() {
    int n;
    scanf("%d", &n);
    int highIdx = 0, lowIdx = 0;
    for (int i = 0;i < n;i++) {
        while (getchar() != '\n');
        gets(books[i].name);
        scanf("%lf", &books[i].price);
        if (books[i].price <= books[lowIdx].price) {
            lowIdx = i;
        }
        if (books[i].price >= books[highIdx].price) {
            highIdx = i;
        }
    }
    printf("%.2lf, %s\n%.2lf, %s", books[highIdx].price, books[highIdx].name, books[lowIdx].price,
           books[lowIdx].name);
    return 0;
}

7-2 一帮一

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#include <stdio.h>
struct Student {
    int gender;
    int paired;
    char name[10];
} stu[100];
int main() {
    int n;
    scanf("%d", &n);
    for (int i = 0;i < n;i++) {
        scanf("%d%s", &stu[i].gender, stu[i].name);
    }
    for (int i = 0;i < n;i++) {
        if (!stu[i].paired) {
            stu[i].paired = 1;
            for (int j = n - 1;j >= 0;j--) {
                if (!stu[j].paired && stu[j].gender != stu[i].gender) {
                    stu[j].paired = 1;
                    printf("%s %s\n", stu[i].name, stu[j].name);
                    break;
                }
            }
        }
    }
    return 0;
}

7-3 计算职工工资

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#include <stdio.h>
int main() {
    int n;
    scanf("%d", &n);
    char name[256];
    double a, b, c;
    for (int i = 0;i < n;i++) {
        scanf("%s%lf%lf%lf", name, &a, &b, &c);
        printf("%s %.2lf\n", name, a + b - c);
    }
    return 0;
}

实验1 顺序结构

7-1 逆序的三位数

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#include<stdio.h>
int main() {
    int a, t = 0;
    scanf("%d", &a);
    while (a) {
        t = t * 10 + a % 10;
        a /= 10;
    }
    printf("%d", t);
    return 0;
}

7-2 求整数均值

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#include<stdio.h>
#define ll long long
int main() {
    int a;
    ll sum = 0;
    for (int i = 0;i < 4;i++) {
        scanf("%d", &a);
        sum += a;
    }
    printf("Sum = %lld; Average = %.1lf", sum, sum / 4.0);
    return 0;
}

7-3 日期格式化

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#include<stdio.h>
int main() {
    int m, d, y;
    scanf("%d-%d-%d", &m, &d, &y);
    printf("%04d-%02d-%02d", y, m ,d);
    return 0;
}

7-4 混合类型数据格式化输入

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#include<stdio.h>
int main() {
    float v1, v4;
    int v2;
    char v3;
    scanf("%f %d %c %f", &v1, &v2, &v3, &v4);
    printf("%c %d %.2f %.2f", v3, v2, v1, v4);
    return 0;
}

7-5 然后是几点

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#include<stdio.h>
int main() {
    int a,b;
    scanf("%d%d", &a, &b);
    int m = (a / 100) * 60 + a % 100;
    m += b;
    while (m < 0) {
        m += 60 * 24;
    }
    int h = m / 60;
    m %= 60;
    h %= 24;
    printf("%d%02d", h, m);
    return 0;
}

7-6 计算存款利息

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#include<stdio.h>
#include<math.h>
int main() {
    double m, y, r;
    scanf("%lf%lf%lf", &m, &y, &r);
    double v = m * pow(1 + r, y) - m;
    printf("interest = %.2lf", v);
    return 0;
}

7-7 苹果装盘

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#include<stdio.h>
int main() {
    int a;
    scanf("%d", &a);
    printf("%d", a / 3 + (a % 3 != 0));
    return 0;
}

7-8 计算并联电阻的阻值

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#include<stdio.h>
int main() {
    double r1, r2;
    scanf("%lf%lf", &r1, &r2);
    printf("%.2lf", 1.0 / (1.0/r1 + 1.0/r2));
    return 0;
}