HDU-2016 Multi-University Training Contest 3-Sqrt Bo-大數開方

kewlgrl發表於2016-07-26
AI

Sqrt Bo

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0


Problem Description
Let's define the function f(n)=⌊n−−√⌋.

Bo wanted to know the minimum number y which satisfies fy(n)=1.

note:f1(n)=f(n),fy(n)=f(fy−1(n))

It is a pity that Bo can only use 1 unit of time to calculate this function each time.

And Bo is impatient, he cannot stand waiting for longer than 5 units of time.

So Bo wants to know if he can solve this problem in 5 units of time.
 

Input
This problem has multi test cases(no more than 120).

Each test case contains a non-negative integer n(n<10100).
 

Output
For each test case print a integer - the answer y or a string "TAT" - Bo can't solve this problem.
 

Sample Input
233
233333333333333333333333333333333333333333333333333333333





 



Sample Output




3
TAT







 



題目意思：



對一個數開方，能在五次之內取整得到0的就輸出次數；不能或者是0的話就輸出TAT。。







解題思路：



這題longlong暴力就能過，隊友很快就過了~然而我當時沒想到，只想到了大數……QwQ



吭哧吭哧改了半天……才過…╮(╯▽╰)╭



longlong的程式碼很簡單我就不寫了，下面給出我寫的大數的…這個模板說不定以後用的上呢~2333








#include <iostream>
using namespace std;
#include <stdlib.h>
#include <string.h>
#define MAXN 1000000
int j,l,size,num,x[MAXN];
char ch[MAXN],ans[MAXN];
void sqrt(char *str)//大數求平方
{
    double i,r,n;
    size=strlen(str);
    if (size%2==1)
    {
        n=str[0]-48;
        l=-1;
    }
    else
    {
        n=(str[0]-48)*10+str[1]-48;
        l=0;
    }
    r=0,num=0;
    while (true)
    {
        i=0;
        while (i*(i+20*r)<=n)
            ++i;
        --i;
        n-=i*(i+20*r);
        r=r*10+i;
        x[num]=(int)i;
        ++num;
        l+=2;
        if (l>=size)
            break;
        n=n*100+(double)(str[l]-48)*10+(double)(str[l+1]-48);
    }


}
int main()
{
    while (cin>>ch)
    {
        int flag=0;
        sqrt(ch);
        if((x[num-1]==1)&&num==1)//注意判斷條件
        {
            cout<<"1"<<endl;
            flag=1;
            goto A;
        }
        for(int i=2; i<=5; ++i)//最多計算五次
        {
            for (j=0; j<num; ++j)
                ans[j]='0'+x[j];//int轉char
            ans[num]='\0';//注意末尾別忘了加結束符
            //cout<<ans<<endl;
            sqrt(ans);//開方
          if((x[num-1]==1)&&num==1)
            {
                cout<<i<<endl;//輸出次數
                flag=1;//標記
                break;
            }
        }
       A: if(!flag)
            cout<<"TAT"<<endl;//超過五次或者為0時
    }
    return 0;
}




            




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