【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+200;
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[1] = false ;
    int tot = 0 ;
    for(ll i=2;i<N;i++){
        if( Is_prime[i] ){
            prime[tot++] = i;
            for(int j=i*2;j<N;j+=i){
                Is_prime[j] = false;
            }
        }
    }
    cnt = tot;
}
int F(int x){
    int res = 0 ;
    while( x ){
        res += (x%10)*(x%10);
        x /= 10;
    }
    return res ;
}
int dfs(int x){
    //printf("%d —— 
",x);
    if( x == 1 ) return dp[x] = 1 ;
    if( vis[x] ) return dp[x] ;
    if( vis[x] == 0 ){
        vis[x] = 1;
        return dp[x] = dfs(F(x));
    }
}
void Mark(int x){
    if( x == 1 ) return ;
    dp[x] = 1 ;
    Mark( F(x) );
}
int main()
{
 
    Euler();
    /*
    for(int i=0;i<10;i++){
        printf("%d
",prime[i]);
    }
    */
    int T,kase,n;
    scanf("%d",&T);
    vis[1] = dp[1] = 1 ;
    while(T--){
        scanf("%d%d",&kase,&n);
        if( Is_prime[n] ){
            int t = dfs(n);
            if( t ){
                Mark(n);
                printf("%d %d YES
",kase,n);
            }else{
                printf("%d %d NO
",kase,n);
            }
        }else{
            printf("%d %d NO
",kase,n);
        }
    }
    return 0 ;
}