1 solutions
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0
C :
#include <stdio.h> int main() { int i; float m=0,n=0,sum=0; scanf("%f %f",&m,&n); for(i=1;i<=n;i++) { sum+=(m+m/2); m=m/2; } printf("%.2f %.2f\n",m,sum-m); return 0; }C++ :
#include<stdio.h> int main() { float sum,m; int n; scanf("%f%d",&m,&n); sum=m; while(n--) { m=m/2; sum+=2*m; } printf("%.2f %.2f\n",m,sum-2*m); }
- 1
Information
- ID
- 19328
- Time
- 1000ms
- Memory
- 128MiB
- Difficulty
- (None)
- Tags
- # Submissions
- 0
- Accepted
- 0
- Uploaded By