标签:number put preview ann 存在 ini stand while cas
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2222 Accepted Submission(s): 883
#include<bits/stdc++.h> using namespace std; const int N=1e7+10; typedef long long ll; char a[N]; int main(){ int len,ans; ll num,cnt; while(~scanf("%s",&a)){ len=strlen(a); if(len>10) printf("TAT\n"); else{ num=0; for(int i=0;i<len;i++){ //将数组里的数变成要求的数 num=num*10+a[i]-‘0‘; } if(num>=4013729316||num==0) printf("TAT\n"); else{ ans=0; while(num!=1){ num=(ll)sqrt(num*1.0); //因为向下取整,所以强制转换一下 ans++; } printf("%d\n",ans); } } } return 0; }
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标签:number put preview ann 存在 ini stand while cas
原文地址:http://www.cnblogs.com/ZERO-/p/7154793.html