约数个数定理及实现

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约束个数定理：
对于一个大于1正整数n可以分解质因数：n=p1^k1*p2^k2*p3^k3.......
则约数个数x=(k1+1)*(k2+1)*(k3+1)*......
1：给你一个n求出最小的约束为n个的数。

思路：建树搜索，以每一个pi为一层建树搜索。

#include<stdio.h>#include<algorithm>#include<string.h>using namespace std;#define ll long longconst ll maxm = 1e18;ll ans, n;int prime[10] = { 2,3,5,7,11,13,17,19,23,29 };void dfs(int a, int b, ll temp);int main(){while (scanf("%d", &n) != EOF){ans = maxm;dfs(1, 0, 1);printf("%I64d\n", ans);}return 0;}void dfs(int a, int b, ll temp){if (a == n&&ans >= temp)ans = temp;if (a >= n)return;for (int i = 1;i <= 64;i++){if (ans / prime[b] < temp)break;temp *= prime[b];dfs(a*(i + 1), b + 1, temp);}}

2:求n以内约数个数最多的数，如果有多个输出最小值。
基本和上题一样。

#include<stdio.h>#include<string.h>#include<algorithm>#include<iostream>using namespace std;#define ll long longconst int ll maxm = ~0ULL;ll n, ans, sum = 0;int p[16] = { 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53 };void dfs(int a, int b, ll temp);int main(){while (cin >> n){ans = maxm;sum = 0;dfs(1, 0, 1);cout << ans << endl;}return 0;}void dfs(int a, int b, ll temp){if (b >= 16)return;if (a >= sum){if (a > sum){sum = a;ans = temp;}else if (a == sum&&temp < ans)ans = temp;}for (int i = 1;i <= 64;i++){if (temp*p[b] > n)break;temp *= p[b];dfs(a*(i + 1), b + 1, temp);}}

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