Water's coin
Time Limit : 10000/5000ms (Java/Other) Memory Limit : 65535/32768K (Java/Other)
Total Submission(s) : 452 Accepted Submission(s) : 72
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Problem Description
Mr. Water find a interesting machine in street. There is a number n showed on the screen. When he put x coins in it, the machine will return y coins to him. And after trying several times , Mr. Water find the number of y is equal to the front n digits of x*x。Mr. Water want to know how much he can get at most, can you tell him?
Notes: Water is a doubi, every time he will put all his coins in machine.
Input
Input contains an integer T(1<=T<=200) in the first line, and then T lines follow. Each line consists of a pair of integers n(1<=n<=9)and x(0<=x<10^n), separated by a space, one pair of integers per line.
Output
For each pair of input integers a and b you should output the max number of coins Mr. Water can get.
The first test
6->36
3->9
9->81
8->64
6->36
Sample Input
2 1 6 2 9
Sample Output
9 81
//题意：
给你一个n，表示数的前n位数，再给一个数x,表示初始的值，问一直将x平方，并每次取它的前n位数，问最大的那个数是什么？
//思路：
大白书上的一道题，貌似是Floyd判环问题（这个还不太懂），直接看代码吧，代码好理解
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define INF 0x3f3f3f3f
#define N 10010
#define ll long long
using namespace std;
int a[110];
int next(int n,int k)
{
if(!k)
return 0;
ll k2=(ll)k*k;
int l=0;
while(k2>0)
{
a[l++]=k2%10;
k2/=10;
}
if(n>l)
n=l;
int ans=0;
for(int i=0;i<n;i++)
ans=ans*10+a[--l];
return ans;
}
int main()
{
int t,n,i,j,k;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&k);
int ans=k;
int k1=k;
int k2=k;
while(1)
{
k1=next(n,k1);
k2=next(n,k2);
if(k2>ans)
ans=k2;
k2=next(n,k2);
if(k2>ans)
ans=k2;
if(k1==k2)//追上以后停止
break;
}
printf("%d\n",ans);
}
return 0;
}
