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

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 long
using 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;
}