题目链接：https://www.51nod.com/onlineJudge/questionCode.html#!problemId=1060

题解：可以去学习一下反素数。

#include <iostream>#include <cstring>#define inf 1000000000000000007using namespace std;typedef unsigned long long ull;const int M = 1e6 + 10;ull n , dp[M];int prime[16] = {2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53};void dfs(int deep , ull sum , int num) {    dp[num] = min(dp[num] , sum);    for(int i = 1 ; i <= 63 ; i++) {        if(sum > 1e18 / prime[deep]) break;        dfs(deep + 1 , sum * prime[deep] , num * (i + 1));        sum *= prime[deep];    }}int main() {    int t;    scanf("%d" , &t);    for(int i = 0 ; i < M ; i++) dp[i] = -inf;    dfs(0 , 1 , 1);    while(t--) {        scanf("%lld" , &n);        int ans;        for(int i = M - 1 ; i >= 1 ; i--) {            if(dp[i] <= n && dp[i] != 0) {ans = i;  break;}        }        printf("%lld %d\n" , dp[ans] , ans);    }    return 0;}

51nod 1060 最复杂的数（数论，反素数）