codeforces 1025B Weakened Common Divisor(质因数分解)

题意：

给你n对数，求一个数，可以让他整除每一对数的其中一个

思路：

枚举第一对数的质因数，然后暴力

代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
#include<functional>

#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
#define lowbit(x) ((x)&(-x)) 

using namespace std;

typedef double db;
typedef long double ldb;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;

const db eps = 1e-;
const int mod = 1e9+;
const int maxn = 2e7+;
const int maxm = 2e6+;
const int inf = 0x3f3f3f3f;
const db pi = acos(-1.0);

inline int read(){
    int num;
    char ch;
    while((ch=getchar())<'' || ch>'');
    num=ch-'';
    while((ch=getchar())>='' && ch<=''){
        num=num*+ch-'';
    }
    return num;
}
int c = ;
int p[maxn];
void d(int x){
    for(int i=;1ll*i*i<=x;i++)if(x%i==){
        p[c++]=i;
        while(x%i==)x/=i;
    }
    if(x>)p[c++]=x;
}
ll gcd(ll a, ll b){
    return b ==  ? a : gcd(b, a % b);
}
PLL pa[ + ];
bool cmp(PLL a, PLL b){
    return max(a.fst, a.sc) < max(b.fst, b.sc);
}
int main() {
    int n;
    scanf("%d", &n);

    for(int i = ; i < n; i++){
        scanf("%I64d %I64d", &pa[i].fst, &pa[i].sc);
    }
    //sort(pa, pa+n, cmp);
    d(pa[].fst);
    d(pa[].sc);
    if(n==){
        printf("%I64d", pa[].fst);
        return ;
    }
    for(int i = ; i < c; i++){
        int flg = ;
        for(int j = ; j < n && flg; j++){
            if(pa[j].fst%p[i]!= && pa[j].sc%p[i]!=) flg = ;

        }
        if(flg){
            printf("%d", p[i]);
            return ;
        }
    }
    printf("-1");
    return ;
}