51Nod1179 最大的最大公约数

Problem

给出N个正整数，找出N个数两两之间最大公约数的最大值。例如：N = 4，4个数为：9 15 25 16，两两之间最大公约数的最大值是15同25的最大公约数5。

Solution

把每个数所有因数搞出来，如果第二次出现就停止。

Code

#include<stdio.h>
#include<set>
#include<iostream>
#include<stack>
#include<cstring>
#include<cmath>
#include<vector>
#include<algorithm>
typedef long long ll;
typedef long double ld;
typedef double db;
#define io_opt ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
using namespace std;
const int mod=998244353;
int mo(ll a,int p){
    return a>=p?a%p:a;
}
inline int rd() {
    int x = 0, f = 1;
    char ch;
    while (ch < '0' || ch > '9') {
        if (ch == '-')f = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return f * x;
}
int cnt=168;
int pri[200]={0,2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,
        53,59,61,67,71,73,79,83,89,97,
        101,103,107,109,113,127,131,137,139,149,
        151,157,163,167,173,179,181,191,193,197,199,
        211,223,227,229,233,239,241,
        251,257,263,269,271,277,281,283,293,
        307,311,313,317,331,337,347,349,
        353,359,367,373,379,383,389,397,
        401,409,419,421,431,433,439,443,449,
        457,461,463,467,479,487,491,499,
        503,509,521,523,541,547,
        557,563,569,571,577,587,593,599,
        601,607,613,617,619,631,641,643,647,
        653,659,661,673,677,683,691,
        701,709,719,727,733,739,743,
        751,757,761,769,773,787,797,
        809,811,821,823,827,829,839,
        853,857,859,863,877,881,883,887,
        907,911,919,929,937,941,947,
        953,967,971,977,983,991,997};

int a[50020][30];
int ct[50020][30];
int n,tmp,mx;
int num[50020];
int ans[1000020];
inline int speed(int a,int b){
    int cur=a,anss=1;
    while(b){
        if(b&1) anss=anss*cur;
        cur=cur*cur;
        b>>=1;
    }
    return anss;
}
void dfs(int t,int cur,int anx){
    if(cur==a[t][0]+1){
        ans[anx]++;
        if(ans[anx]>1) mx=max(mx,anx);
        return;
    }
    for(int i=0;i<=ct[t][cur];i++){
        dfs(t,cur+1,anx*speed(a[t][cur],i));
    }
}
int main(){
    //io_opt;
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%d",&num[i]);
    }
    sort(num+1,num+1+n);
    for(int i=n;i>=1;i--){
        tmp=num[i];
        for(int j=1;j<=cnt;j++){
            if(num[i]%pri[j]==0){
                a[i][++a[i][0]]=pri[j];
            }
            while(num[i]%pri[j]==0){
                ct[i][a[i][0]]++;
                num[i]/=pri[j];
            }
        }
        if(num[i]>1){
            a[i][++a[i][0]]=num[i];
            ct[i][a[i][0]]++;
        }
        if(mx>=tmp){
            break;
        }dfs(i,1,1);
    }
    printf("%d
",mx);

    return 0;
}