1 条题解
-
0
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