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 = 10100000;
ll n;
stack <int> s;

void init_prim(){
    memset(visit, true, sizeof(visit));
    int num = 0;
    for (int i = 2; i <= nn; ++i){
        if (visit[i] == true){
            num++;
            prime[num] = i;
        }
        for (int j = 1; ((j <= num) && (i * prime[j] <= nn));  ++j){
            visit[i * prime[j]] = false;
            if (i % prime[j] == 0) break; //点睛之笔
        }
    }
}

ll quickpow(ll m,ll n,ll k){
    int b = 1;
    while (n > 0){
          if (n & 1)
             b = (b*m)%k;
          n = n >> 1 ;
          m = (m*m)%k;
    }
    return b;
}

ll getsum(int x){
    int sum = 0;
    while(x){
        sum += x % 10;
        x = x / 10;
    }
    return sum;
}

bool witness(ll a,ll n){
    ll t,d,x;
    d = 1;
    int i=ceil(log(n-1.0)/log(2.0)) - 1;//j
    for(;i>=0;i--)
    {
        x=d;  d=(d*d)%n;
        if(d==1 && x!=1 && x!=n-1) return true;
        if( ((n-1) & (1<<i)) > 0)
            d=(d*a)%n;
    }
    return d==1? false : true;
}
bool miller_rabin(ll n){
    int s[]={2,7,61};
    if(n==2 || n == 1)    return true;
    if(n==1 || ((n&1)==0))    return false;
    for(int i=0;i<3;i++)// 3
        if(witness(s[i], n))    return false;
    return true;
}

bool isPrime(ll n){
    if(n == 1 || n == 2 || n == 3 || n == 5) return true;
    else if(n % 2 == 0 || n % 3 == 0 || n % 5 == 0)  return false;
    for(int i = 2; i <= sqrt(n); ++i){
        if(n % i == 0) return false;
    }
    return true;
}

bool judge(ll x){
    int sum1, sum2 = 0, i;
    sum1 = getsum(x);
    for(i = 2; i <= sqrt(x); ++i){
        if(x % i == 0){
            s.push(i);
            x = x / i;
            while(x % i == 0){
                s.push(i);
                x = x / i;
            }
        }
        if(x == 1)
            break;
    }
    if(x > 1) 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 <= 0)  break;
        ll num = n;
        while(1){
            ++num;
            if(isPrime(num))    continue;
            else if(judge(num)){
                cout << num << endl;
                break;
            }
        }
    }
    return 0;
}