吼吼哈嘿

统计类型

给出一个字符串，输出出现次数最多的三个字母，次数相同按照ASCII码由小到大排列，不缺分大小写，输出为小写字母。

#include<stdio.h>
int main(){
    char string[1005] = {0};
    int counts[26] = {0};
    scanf("%s", string);
    for (int i = 0; string[i] != '\0'; i++) {
        if ('a' <= string[i] <= 'z')
        {
            counts[string[i] - 'a']++;
        }
        else if ('A' <= string[i] <= 'Z')
        {
            counts[string[i] - 'A']++;
        }
    }
    int maxCount = -1;
    int maxNumber = -1;
    for (int i = 0; i < 26; i++)
    {
        if (maxCount < counts[i])
        {
            maxCount = counts[i];
            maxNumber = i;
        }
    }
    printf("%c\n", maxNumber + 'a');
    return 0;
}

CCF中出现的类似题目：

#include<stdio.h>
int main(){
    int counts[1001] = {0};
    int numberOfDate = 0;
    int oneDate = 0;
    scanf("%d", &numberOfDate);
    for (int i = 0; i < numberOfDate; i++) {
        scanf("%d", &oneDate);
        counts[oneDate]++;
    }
    int max = -1;
    int theMostNumber = -1;
    for (int i = 0; i < 1001; i++) {
        if (counts[i] > max) {
            max = counts[i];
            theMostNumber = i;
        }
    }
    printf("%d\n", theMostNumber);
    return 0;
}

数学类型

求1-n之间的质数

#include<stdio.h>
int main(){
    int n = 0;
    scanf("%d", &n);
    for (int i = 2; i < n; i++) {
        bool isZhiShu = true;
        for (int j = 2; j <= i / 2; j++) {
            if (i % j == 0) {
                isZhiShu = false;
                break;
            }
        }
        if (isZhiShu)
        {
            printf("%d ", i);
        }
    }
    printf("\n");
    return 0;
}

求a,b两个数的最小公约数 最大公倍数

#include<stdio.h>
// 辗转相除法
int gcd(int a, int b) {
    if (!b) {
        return a;
    } else {
        return gcd(b, a%b);
    }
}
int main(){
    int a = 0;
    int b = 0;
    scanf("%d %d", &a, &b);
    // 最大公约数
    int maxCommon = -1;
    if (a > b) {
        maxCommon = gcd(a, b);
        printf("%d\n", maxCommon);
    } else {
        maxCommon = gcd(b, a);
        printf("%d\n", maxCommon);
    }
    // 最小公倍数
    printf("%d\n", a * b / maxCommon);
    return 0;
}

水仙花数是指一个 n 位数 ( n≥3 )，它的每个位上的数字的 n 次幂之和等于它本身。（例如：1^3 + 5^3+ 3^3 = 153）
求100 - 999的水仙花数

#include<stdio.h>
int nnn(int n) {
    return n * n * n;
}
int main(){
    for (int i = 100; i <= 999; i++) {
        int n3 = i % 10;
        int n2 = i /10 % 10;
        int n1 = i / 100;
        int n = nnn(n1) + nnn(n2) + nnn(n3);
        if (n == i)
        {
            printf("%d ", n);
        }
    }
    printf("\n");
    return 0;
}