ural 1807

给定一个数，分解成k个数之和，使得这K个数公约数最大，最小公倍数也最大。

最大公约数最大，肯定选择除以素数分解后的最小素数。

现在相当于分解质数m使得最小公倍数最大。

用dp[i][j]表示用前i个质数分解和为m最小公倍数最大。

最后一定是2 ^ k1 * 3 ^ k2 * 5 ^ k3....

所以可以写出dp转移出来。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>

using namespace std;

const int INF = 1000000007;
const int maxn = 100007;
long long least;
int vis[maxn], prime[maxn], prime_num;
int ans[maxn], ans_num, temp[maxn];
int pre[156][35000];
double dp[156][35000];
long long big_lcm;
long long n, f;

void getAllPrime() {
    memset(vis, 0, sizeof(vis));
    for (int i = 2; i <= sqrt(maxn * 1.0); i++) {
        if (!vis[i]) {
            for (int j = i * i; j < maxn; j += i) {
                vis[j] = 1;
            }
        }
    }
    prime_num = 0;
    for (int i = 2; i < maxn; i++) {
        if (!vis[i]) {
            prime[++prime_num] = i;
        }
    }
}

int main() {
//    freopen("in.txt", "r", stdin);
    long long x;
    getAllPrime();
    scanf("%I64d", &n);
    int flag = 0;
    for (int j = 1; j <= prime_num; j++) {
        if (n % prime[j] == 0) {
            flag = 1;
            least = prime[j];
            break;
        }
    }
    if (!flag) {
        least = n;
    }
    long long big_gcd = n / least;
    big_lcm = 0;
    flag = 0;
    if (least == 2) {
        printf("%I64d %I64d\n", big_gcd, big_gcd);
        return 0;
    }
    if (least == 3) {
        printf("%I64d %I64d\n", big_gcd, big_gcd * 2);
        return 0;
    }
    for (int i = 1; i <= 150; i++) {
        for (int j = 0; j <= least; j++) {
            dp[i][j] = dp[i - 1][j];
            pre[i][j] = j;
            int cnt = prime[i];
            double t = log(prime[i] * 1.0);
            for (int k = 1; cnt <= j; k++) {
                double temp = k * t;
                if (dp[i][j] < dp[i - 1][j - cnt] + temp) {
                    pre[i][j] = j - cnt;
                    dp[i][j] = dp[i - 1][j - cnt] + temp;
                }
                cnt *= prime[i];
            }
        }
    }
//    for (int i = 1; i <= 150; i++) {
//        for (int j = 0; j <= least; j++) {
//            dp[i][j] = dp[i - 1][j];
//            pre[i][j] = j;
//        }
//        for (int j = prime[i]; j <= least; j *= prime[i]) {
//            double temp = log(j * 1.0);
//            for (int k = j; k <= least; k++) {
//                if (dp[i][k] < dp[i - 1][k - j] + temp) {
//                    dp[i][k] = dp[i - 1][k - j] + temp;
//                    pre[i][k] = k - j;
//                }
//            }
//        }
//    }
    int tot = least;
    for (int i = 150; i >= 1; i--) {
        if ((tot - pre[i][tot]) != 0) {
            printf("%I64d ", (tot - pre[i][tot]) * big_gcd);
            tot = pre[i][tot];
        }
    }
    while (tot > 0) {
        printf("%I64d ", big_gcd);
        tot--;
    }
    printf("\n");
    return 0;
}