比赛入口
D Recover it!
做法：排序后从大到小遍历一遍，把合数的情况先输出，再输出第a[i]个质数的情况
代码：
#include<bits/stdc++.h>
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define lson rt<<1, l, mid
#define rson rt<<1|1, mid+1, r
using namespace std;
typedef long long LL;
typedef pair<LL, int> pli;
typedef pair<int, int> pii;
typedef pair<LL, LL> pll;
typedef pair<double, double> pdd;
typedef map<char, int> mci;
typedef map<string, int> msi;
template<class T>
void read(T &res) {
int f = 1; res = 0;
char c = getchar();
while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); }
while(c >= '0' && c <= '9') { res = res * 10 + c - '0'; c = getchar(); }
res *= f;
}
const int N = 3e6;
int prime[N], f[N];
void init() {
for (int i = 2; i < N; i++) {
if(!f[i]) prime[++prime[0]] = i;
for(int j = 1; j <= prime[0] && i * prime[j] <= N; j++) {
f[i * prime[j]] = 1;
if(i % prime[j] == 0) break;
}
}
}
int a[N], b[N];
int main() {
init();
int n; read(n);
for(int i = 0; i < n + n; ++i) {
read(a[i]); ++b[a[i]];
}
sort(a, a+n+n);
int mid;
for(int i = n+n-1; i + 1; --i) {
mid = a[i];
if(b[mid] <= 0 || !f[mid]) continue;
int k = 2;
while(mid % k) ++k;
printf("%d ", mid);
--b[mid]; --b[mid/k];
}
for(int i = 0; i < n + n; ++i) {
if(mid = a[i], b[mid] > 0) {
printf("%d ", mid);
--b[mid]; --b[prime[mid]];
}
}
return 0;
}
F Destroy it!
做法：带限制的01背包问题
