题目描述

好麻烦，繁琐但不是很难，就是情况太多

样例

#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
    int n;
    int a;
    char t;
    int sum=0,sum1=0,sum2=0,sum3=0;//sum1表示兔子,sum2表示老鼠,sum3表示青蛙
    double s1,s2,s3;
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        cin>>a>>t;
        if(t=='C')
            sum1+=a;
        else if(t=='R')
            sum2+=a;
        else
            sum3+=a;
        sum+=a;
    }
    s1=(double)sum1/(double)sum*100;
    s2=(double)sum2/(double)sum*100;
    s3=(double)sum3/(double)sum*100;
    cout<<"Total: "<<sum<<" animals"<<endl;
    cout<<"Total coneys: "<<sum1<<endl;
    cout<<"Total rats: "<<sum2<<endl;
    cout<<"Total frogs: "<<sum3<<endl;
    printf("Percentage of coneys: %.2lf %\n",s1);
    printf("Percentage of rats: %.2lf %\n",s2);
    printf("Percentage of frogs: %.2lf %\n",s3);
    return 0;
}