题目描述

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样例

#include <iostream>
using namespace std;
int main()
{
    int a, n;
    char c;
    scanf("%d", &n);
    int sum = 0, C = 0, R = 0, F = 0; //计数器sum表示总和,C 表示兔子(coney)，R 表示老鼠(rat)，F 表示青蛙(frog)
    while (n -- ) 
    {
        cin>>a>>c;
        if(c=='C') C+=a; 
        if(c=='R') R+=a;
        else if(c=='F') F+=a;
        sum+=a;
    }
    double c2=(double)C/(double)sum*100;
    double r=(double)R/(double)sum*100;
    double f=(double)F/(double)sum*100;
    //这里需要浮点数来存，注意一定要强制类型转换
    printf("Total: %d animals\n", sum);
    printf("Total coneys: %d\n", C);
    printf("Total rats: %d\n", R);
    printf("Total frogs: %d\n", F);
    printf("Percentage of coneys: %.2lf %%\n", c2);
    printf("Percentage of rats: %.2lf %%\n", r);
    printf("Percentage of frogs: %.2lf %%\n", f); 

    return 0;
}

算法1

(暴力枚举) $O(n^2)$

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时间复杂度

参考文献

C++ 代码

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算法2

(暴力枚举) $O(n^2)$

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时间复杂度

参考文献

C++ 代码

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