1 条题解

  • 0
    @ 2024-12-18 14:07:26

    C :

    #include <stdio.h>
    int main() 
    {
    	int n,count,g,sumr=0,sumf=0,sumc=0;
    	char a;
    	scanf("%d",&n);
    	for(count=0;count<n;count++){
    		scanf("%d %c",&g,&a);
    		if(a=='R'){
    			sumr=sumr+g;
    		}
    			if(a=='F'){
    			sumf=sumf+g;
    		}
    			if(a=='C'){
    			sumc=sumc+g;
    		}
    	}
    	printf("Total: %d animals\nTotal coneys: %d\nTotal rats: %d\nTotal frogs: %d\nPercentage of coneys: %.2f %%\nPercentage of rats: %.2f %%\nPercentage of frogs: %.2f %%\n",sumc+sumf+sumr,sumc,sumr,sumf,(float)sumc/(sumc+sumf+sumr)*100.0,(float)sumr/(sumc+sumf+sumr)*100.0,(float)sumf/(sumc+sumf+sumr)*100.0);
    	return 0;
    }
    

    C++ :

    #include<cstdio>
    using namespace std;
    
    int main()
    {
        int n,N,a=0,b=0,c=0;
        char T;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
    
        {
    
          scanf("%d %c",&N,&T);
    
          if(T=='C') a=a+N;
    
          else if(T=='R') b=b+N;
    
          else c=c+N;
        }
    
    
        printf("Total: %d animals\n",a+b+c);
        printf("Total coneys: %d\n",a);
        printf("Total rats: %d\n",b);
        printf("Total frogs: %d\n",c);
        printf("Percentage of coneys: %.2lf %%\n",(double)a/(a+b+c)*100);
        printf("Percentage of rats: %.2lf %%\n",(double)b/(a+b+c)*100);
        printf("Percentage of frogs: %.2lf %%\n",(double)c/(a+b+c)*100);
    
    
        return 0;
    }
    
    • 1

    信息

    ID
    619
    时间
    1000ms
    内存
    128MiB
    难度
    7
    标签
    提交数
    125
    已通过
    30
    上传者