2016 Multi-University Training Contest 3 1001 Sqrt Bo (模拟)
2016-08-01 09:37
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题目链接
简单题意
给出一个数,如果能在5次以内的开方并向下取整后变为一,则输出操作次数,否则输出“TAT”思路
很明显,不同次数的操作变成一的数是有上下界的,找出来,然后直接和输入比较就可以了,注意5次操作的上界恰好是int的上界,所以最好用longlong。代码
#include <bits/stdc++.h> using namespace std; long long lim[10] = {0,4,16,256,65536,4294967296}; string str; int main(){ while(cin >> str){ if(str.length() > 10) {puts("TAT");continue;} long long tmp = atoll(str.c_str()); if(tmp >= lim[5]) {puts("TAT");continue;} int ans = 0; if(tmp == 0 ){puts("TAT");continue;} for(int i = 0 ; i < 5 ; i ++){ if(tmp >= lim[i]) ans = i+1; } printf("%d\n",ans); } }
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