POJ 1142 质因数分解

只要很朴素的分解就可以了，数据量不大

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <iostream>
#include <stack>
#include <algorithm>
 
#define ll long long
using namespace std;
const int INF = 0x3f3f3f3f;
const int MAXSIZE = ;
ll n;
stack <int> s;
 
void init_prim(){
    memset(visit, true, sizeof(visit));
    int num = ;
    for (int i = ; i <= nn; ++i){
        if (visit[i] == true){
            num++;
            prime[num] = i;
        }
        for (int j = ; ((j <= num) && (i * prime[j] <= nn));  ++j){
            visit[i * prime[j]] = false;
            if (i % prime[j] == ) break; //点睛之笔
        }
    }
}
 
ll quickpow(ll m,ll n,ll k){
    int b = ;
    while (n > ){
          if (n & )
             b = (b*m)%k;
          n = n >>  ;
          m = (m*m)%k;
    }
    return b;
}
 
ll getsum(int x){
    int sum = ;
    while(x){
        sum += x % ;
        x = x / ;
    }
    return sum;
}
 
bool witness(ll a,ll n){
    ll t,d,x;
    d = ;
    int i=ceil(log(n-1.0)/log(2.0)) - ;//j
    for(;i>=;i--)
    {
        x=d;  d=(d*d)%n;
        if(d== && x!= && x!=n-) return true;
        if( ((n-) & (<<i)) > )
            d=(d*a)%n;
    }
    return d==? false : true;
}
bool miller_rabin(ll n){
    int s[]={,,};
    if(n== || n == )    return true;
    if(n== || ((n&)==))    return false;
    for(int i=;i<;i++)//
        if(witness(s[i], n))    return false;
    return true;
}
 
bool isPrime(ll n){
    if(n ==  || n ==  || n ==  || n == ) return true;
    else if(n %  ==  || n %  ==  || n %  == )  return false;
    for(int i = ; i <= sqrt(n); ++i){
        if(n % i == ) return false;
    }
    return true;
}
 
bool judge(ll x){
    int sum1, sum2 = , i;
    sum1 = getsum(x);
    for(i = ; i <= sqrt(x); ++i){
        if(x % i == ){
            s.push(i);
            x = x / i;
            while(x % i == ){
                s.push(i);
                x = x / i;
            }
        }
        if(x == )
            break;
    }
    if(x > ) s.push(x);
    while(!s.empty()){
        sum2 += getsum(s.top());
        s.pop();
    }
    if(sum1==sum2) return true;
    else return false;
}
int main(){
    int i, j, k;
    while(cin >> n){
        if(n <= )  break;
        ll num = n;
        while(){
            ++num;
            if(isPrime(num))    continue;
            else if(judge(num)){
                cout << num << endl;
                break;
            }
        }
    }
    return ;
}