【记忆化搜索】Happy Happy Prime Prime

题目描述

RILEY VASHTEE: [reading from display] Find the next number in the sequence:
313 331 367 ...? What?
THE DOCTOR: 379.
MARTHA JONES: What?
THE DOCTOR: It’s a sequence of happy primes — 379.
MARTHA JONES: Happy what?
THE DOCTOR: Any number that reduces to one when you take the sum of the square of its digits and continue iterating it until it yields 1 is a happy number. Any number that doesn’t, isn’t. A happy prime is both happy and prime.
THE DOCTOR: I dunno, talk about dumbing down. Don’t they teach recreational mathematics anymore?
Excerpted from “Dr. Who”, Episode 42 (2007).
The number 7 is certainly prime. But is it happy?

                                        

It is happy :-). As it happens, 7 is the smallest happy prime. Please note that for the purposes of this problem, 1 is not prime.
For this problem you will write a program to determine if a number is a happy prime.

输入

The first line of input contains a single integer P, (1 ≤ P ≤ 1000), which is the number of data sets that follow. Each data set should be processed identically and independently.
Each data set consists of a single line of input. It contains the data set number, K, followed by the happy prime candidate, m, (1 ≤ m ≤ 10000).

输出

For each data set there is a single line of output. The single output line consists of the data set number,K, followed by a single space followed by the candidate, m, followed by a single space, followed by ‘YES’or ‘NO’, indicating whether m is a happy prime.

样例输入

4
1 1
2 7
3 383
4 1000

样例输出

1 1 NO
2 7 YES
3 383 YES
4 1000 NO

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【题解】

分析一下，其实最多也就5位，每个位置上都是9，也就是81*5=405.也就是说每次搜索记忆化搜索就可。

具体看代码就懂了。

 #include<bits/stdc++.h>
 using namespace std;
 const int N = 1e4+;
 typedef long long ll;
 int prime[N],cnt;
 int dp[N],vis[N];
 bool Is_prime[N];
 void Euler(){
     memset( Is_prime , true , sizeof Is_prime );
     Is_prime[] = false ;
     int tot =  ;
     for(ll i=;i<N;i++){
         if( Is_prime[i] ){
             prime[tot++] = i;
             for(int j=i*;j<N;j+=i){
                 Is_prime[j] = false;
             }
         }
     }
     cnt = tot;
 }
 int F(int x){
     int res =  ;
     while( x ){
         res += (x%)*(x%);
         x /= ;
     }
     return res ;
 }
 int dfs(int x){
     //printf("%d —— \n",x);
     if( x ==  ) return dp[x] =  ;
     if( vis[x] ) return dp[x] ;
     if( vis[x] ==  ){
         vis[x] = ;
         return dp[x] = dfs(F(x));
     }
 }
 void Mark(int x){
     if( x ==  ) return ;
     dp[x] =  ;
     Mark( F(x) );
 }
 int main()
 {

     Euler();
     /*
     for(int i=0;i<10;i++){
         printf("%d\n",prime[i]);
     }
     */
     int T,kase,n;
     scanf("%d",&T);
     vis[] = dp[] =  ;
     while(T--){
         scanf("%d%d",&kase,&n);
         if( Is_prime[n] ){
             int t = dfs(n);
             if( t ){
                 Mark(n);
                 printf("%d %d YES\n",kase,n);
             }else{
                 printf("%d %d NO\n",kase,n);
             }
         }else{
             printf("%d %d NO\n",kase,n);
         }
     }
     return  ;
 }