例题3-5，例题4-2，例题4-3

Posted 2020-11-10 ton2018

例题3-5生成元

 1 #include <stdio.h>
 2 #include <string.h>
 3 #define maxn 100005
 4 int ans[maxn];
 5 
 6 int main() {
 7     int T, n;
 8     memset(ans, 0, sizeof(ans));
 9     for (int m = 1; m < maxn; m++) {
10         int x = m, y = m;
11         while (x > 0) {
12             y += x % 10;
13             y /= 10;
14         }
15         if(ans[y] == 0 || m < ans[y])
16             ans[y] = m;
17     }
18     scanf("%d", &T);
19     while(T--) {
20         scanf("%d", &n);
21         printf("%d\n", ans[n]);
22     }
23     return 0;
24 }

例题4-2 刽子手游戏

#include <stdio.h>
#include <string.h>
#define maxn 100
int left, chance;   //还需猜测left个位置，错chance次之后就输了
char s[maxn], s2[maxn];     //答案是字符串s，玩家猜的字母序列是s2
int win, lose;      //win=1表示已经赢了；lose=1表示已经输了

void guess(char ch) {
    int bad = 1;
    for(int i = 0; i < strlen(s); i++) {
        if(s[i] == ch) {
            left--;
            s[i] = ‘ ‘;
            bad = 0;
        }
        if(bad)
            --chance;
        if(!chance)
            lose = 1;
        if(!left)
            win = 1;
    }
}
int main() {
    int rnd;
    while (scanf("%d%s%s",&rnd, s, s2) == 3 && rnd != -1) {
        printf("Round %d\n", rnd);
        win = lose = 0;     //求解一组新数据之前要初始化
        left = strlen(s);
        chance = 7;
        for (int i = 0; i < strlen(s2); i++) {
            guess(s2[i]);       //猜测一个字母
            if(win || lose)     //检查状态
                break;
        }
        if(win)
            printf("You win.\n");
        else if(lose)
            printf("You lose.\n");
        else
            printf("You chickened out.\n");
    }
    return 0;
}

例题4-3救济金发放

#include <stdio.h>
#define maxn 25
int n, k, m, a[maxn];

//逆时针走t步，步长是d（-1表示顺时针走），返回新位置
int go(int p, int d, int t) {
    while (t--) {
        do {
            p = (p+d+n-1) %n + 1;
        }while(a[p] == 0);      //走到下一个非0数字
    }
    return p;
}

int main() {
    while (scanf("%d%d%d", &n, &k, &m) == 3 && n) {
        for (int i = 1; i <= n; i++)
            a[i] = i;
        int left = n;
        int p1 = n, p2 = 1;
        while(left) {
            p1 = go(p1, 1, k);
            p2 = go(p2, -1, m);
            printf("%3d", p1);
            left--;
            if(p2 != p1) {
                printf("%3d", p2);
                left--;
            }
            a[p1] = a[p2] = 0;
            if (left)
                printf(",");
        }
        printf("\n");
    }
    return 0;
}