2017 Multi-University Training Contest - Team 1 Add More Zero
2017-07-28 20:36
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http://acm.hdu.edu.cn/showproblem.php?pid=6033
log102可以用换底公式(log(2) / (log(10)替换,最后答案向下取整。
题意:
跟你一个m,求一个k的最小值,使得10k>2m - 1思路:
其实就是求n * log102,非常简单。log102可以用换底公式(log(2) / (log(10)替换,最后答案向下取整。
代码:
#include<bits/stdc++.h> using namespace std; int main(){ int n; int cas = 1; while(~scanf("%d",&n)){ double ans = n * (log(2)) / (log(10)); printf("Case #%d: %d\n",cas++,(int)ans); } }
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