hdoj 6033(2017 Multi-University Training Contest - Team 1) Add More Zero
2017-08-02 17:07
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题目链接:Add More Zero
题目大意:给你一个m,现在你可以表示1到2m−1里面的数,能表达到1到10k,问这个k最大是多少
题目思路:实际上就是求2m−1有多少位,也就是问⌊lg(2m−1)⌋是多少,因为不存在2m−1比2m少一位的情况,所以直接求lg(2m)就好了,也就是m∗lg(2),所以这就是答案
题目大意:给你一个m,现在你可以表示1到2m−1里面的数,能表达到1到10k,问这个k最大是多少
题目思路:实际上就是求2m−1有多少位,也就是问⌊lg(2m−1)⌋是多少,因为不存在2m−1比2m少一位的情况,所以直接求lg(2m)就好了,也就是m∗lg(2),所以这就是答案
#include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #include <algorithm> #include <iostream> using namespace std; int main(){ int m,t = 0; while(~scanf("%d",&m)){ printf("Case #%d: ",++t); int ans = m*log10(2.0); printf("%d\n",ans); } return 0; }
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