UVA 11549 模拟 Floyed判圈法的应用 Calculator Conundrum

#include<cstdio>
#include<iostream>
#include<map>
#include<cmath>
using namespace std;
typedef long long LL;
map<int,int> dic;
int wei[10],tot,fans,n,k,T;

inline int Next(int n,int k)
{
LL tmp = (LL)k*k;
int d = log10(tmp)+1;
int ans;
if(d>n)ans = tmp / wei[d-n];
else ans = tmp;
return ans;
}

/*int main()
{
freopen("a.in","r",stdin);
scanf("%d",&T);
wei[1]=10;
for(int i=2;i<=9;i++)wei[i]=wei[i-1]*10;
while(T--)
{
scanf("%d%d",&n,&k);
dic.clear();tot=fans=0;
while(dic[k]==0){
dic[k]=++tot;
if(k>fans)fans=k;
k=Next(n,k);
}
printf("%d\n",fans);
}
return 0;
}*/

//Floyed判圈算法，空间复杂度降到O(1)
//假象有两个小孩在一个有环的跑道上赛跑，其中一个小孩的速度是另一个小孩速度的两倍，跑得快的小孩一定能都追上跑的慢的小孩，根据这一点设计判圈的算法。
int main()
{
freopen("a.in","r",stdin);
scanf("%d",&T);
wei[1]=10;
for(int i=2;i<=9;i++)wei[i]=wei[i-1]*10;
while(T--)
{
scanf("%d%d",&n,&k);
int ans = k;
int k1 = k,k2 = k;
do{
//小孩1每次走一步
k1 = Next(n,k1);
//小孩2每次走两步
k2 = Next(n,k2);if(k2>ans)ans = k2;
k2 = Next(n,k2);if(k2>ans)ans = k2;
}while(k1!=k2); //两小孩相遇，即为圈
printf("%d\n",ans);
}
return 0;
}