1 solutions
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0
C :
#include<stdio.h> #include<math.h> int main() { int y,i; double sum; while(scanf("%d",&y)!=EOF,y) { sum=0; if(y==1960) printf("3\n"); else { for(i=1;;i++) { sum+=log10(i)/log10(2); if(sum>pow(2,(y-1960)/10+2)) break; } printf("%d\n",i-1); } } return 0; }
- 1
Information
- ID
- 18715
- Time
- 1000ms
- Memory
- 128MiB
- Difficulty
- (None)
- Tags
- # Submissions
- 0
- Accepted
- 0
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