hdu5752 2016 Multi-University Training Contest 3 Sqrt Bo 解题报告

来源：互联网发布：python sort函数编辑：程序博客网时间：2024/06/05 11:40

应号召.........又水了一发博客

题意大概是说，给你一个数n<=10^100，问你要多少次向下取整的开根才能使他变成1，如果超过5次就输出TAT

看完觉得挺水的，事实上确实不难

打个表，看哪些值是1-5次向下取整的开根能取到1的极限，然后判一下大小就没问题了

（判大小打错了卡我好久.........毕竟蠢）

#include<set>#include<map>#include<deque>#include<queue>#include<stack>#include<cmath>#include<ctime>#include<vector>#include<string>#include<bitset>#include<cstdio>#include<cstdlib>#include<cstring>#include<climits>#include<complex>#include<iostream>#include<algorithm>#define LL long longusing namespace std;const int maxn = 110000;int zero[maxn] = { 4,2,9,4,9,6,7,2,9,5 },len;LL num[5] = {1,3,15,255,65535};char st[maxn];int count(  ){LL tmp = 0;for( int i=0;i<len;i++ ){( tmp *= 10 ) += st[i]-'0';}if( tmp == 0 ) return 6;for( int i=0;i<5;i++ )if( tmp <= num[i] ) return i;return 5;}int main(){while( scanf("%s",st) != EOF ){len = strlen( st );if( len > 10 ) printf("TAT\n");else if( len < 10 ) {int  tmp = count();if( tmp > 5 ) printf("TAT\n");else printf("%d\n",tmp);}else{bool v = true;for( int i=0;i<len;i++ ){if( st[i]-'0' > zero[i] ){v = false;break;}if( st[i]-'0' < zero[i] ) break;}if( v ){int  tmp = count();if( tmp > 5 ) printf("TAT\n");else printf("%d\n",tmp);}else printf("TAT\n");}}return 0;}

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