【Floyd判圈】计算器谜题

Floyd判圈

传送门：CodeVS4671

Floyd判圈算法：k1、k2同时操作，k1执行1遍，k2执行2遍；由于存在环，k2必定会等于k1

上代码比较简明易懂：

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
#define LL long long
const LL a[10]={1,10,100,1000,10000,100000,1000000,10000000,100000000,1000000000};
LL n,m,k,ans;
LL nxt(LL n,LL x){x=x*x; while (x>=a[n]) x/=10; return x;}
int main(){
	scanf("%lld",&m);
	
	for (;m;m--){
		scanf("%lld%lld",&n,&k);
		LL k1=k,k2=k; ans=k;
		do{
			k1=nxt(n,k1);
			k2=nxt(n,k2); if (k2>ans) ans=k2;
			k2=nxt(n,k2); if (k2>ans) ans=k2;
		}while(k1!=k2);
		printf("%lld\n",ans);
	}
	
	return 0;
}