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水过200~~~~付代码(写的很丑,轻喷)就是2^n<1*2*3...m n<log21+log22+log23.....然后换底公式 #include<stdio.h> #include<math.h> int find(int n) { double t=pow((double)2,(n-1960)/10+2),sum=0; int i=0; for(i=2;sum<t;i++) { sum+=log((double)i)/log((double)2); } return i-2; } int main() { int n=0; while(scanf("%d",&n)&&n) { printf("%d\n",find(n)); } } Followed by: Post your reply here: |
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