題目鏈接

BZOJ2257

題解

由裴蜀定理我們知道，若幹的瓶子如此傾倒最小能湊出的是其\(gcd\)
現在我們需要求出\(n\)個瓶子中選出\(K\)個使\(gcd\)最大
每個數求出因數排序即可

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
 

#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 1005,maxm = 1000005,INF = 1000000000;
inline int read(){
    int out = 0,flag = 1; char c = getchar();
    while (c < 48 || c > 57){if
 
 (c == ‘-‘) flag = -1; c = getchar();}
    while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
    return out * flag;
}
int n,K;
int p[maxm],pi;
void sp(int x){
    p[++pi] = x;
    for (int i = 2; i * i <= x; i++)
        if (x % i == 0){
            p[++pi] = i;
            if
 
 (i * i != x) p[++pi] = x /  i;
        }
}
int main(){
    n = read(); K = read();
    REP(i,n) sp(read());
    sort(p + 1,p + 1 + pi);
    int ans = 1,cnt = 0;
    for (int i = 1; i <= pi; i++){
        if (p[i] != p[i - 1]){
            if (cnt >= K) ans = max(ans,p[i - 1]);
            cnt = 1;
        }
        else cnt++;
    }
    if (cnt >= K) ans = max(ans,p[pi]);
    printf("%d\n",ans);
    return 0;
}

BZOJ2257 [Jsoi2009]瓶子和燃料 【裴蜀定理】