1 solutions
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0
C :
int main(int argc, char* argv[]) { int n,i; double sum,f[100]; while(~scanf("%d",&n)) { f[0]=1;f[1]=2; sum=0; for(i=1;i<=n;i++) { if(i>1) f[i]=f[i-2]+f[i-1]; sum+=f[i]/f[i-1]; } printf("%.2lf\n",sum); } return 0; }C++ :
#include<iostream> #include<cstdio> using namespace std; int main() { int n; cin>>n; double fz=2,fm=1,sum=0; for (int i=1; i<=n; i++) { sum+=fz/fm; double t=fz; fz+=fm; fm=t; } printf("%.2lf\n",sum); return 0; }
- 1
Information
- ID
- 19273
- Time
- 1000ms
- Memory
- 128MiB
- Difficulty
- (None)
- Tags
- (None)
- # Submissions
- 0
- Accepted
- 0
- Uploaded By