Merge pull request #1251(change int to ULL) from 24OI/Ir1d-patch-1

#### 求因子数一定的最小数

题目链接：<http://codeforces.com/problemset/problem/27/E>

题目链接：<https://codeforces.com/problemset/problem/27/E>

对于这种题，我么只要以因子数为 dfs 的返回条件基准，不断更新找到的最小值就可以了

#include <stdio.h>

#define ULL unsigned long long

#define INF ~0ULL

int p[16] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53};

ULL p[16] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53};

ULL ans;

int n;

ULL n;

// depth: 当前在枚举第几个素数。num: 当前因子数。

// temp: 当前因子数量为 num

// 的时候的数值。up：上一个素数的幂，这次应该小于等于这个幂次嘛

void dfs(int depth, int temp, int num, int up) {

void dfs(ULL depth, ULL temp, ULL num, ULL up) {

  if (num > n || depth >= 16) return;

  if (num == n && ans > temp) {

    ans = temp;

}

int main() {

  while (scanf("%d", &n) != EOF) {

  while (scanf("%lld", &n) != EOF) {

    ans = INF;

    dfs(0, 1, 1, 64);

    printf("%d\n", ans);

    printf("%lld\n", ans);

  }

  return 0;

}

}

int main() {

  while (scanf("%lld", &n) != EOF) {

  while (scanf("%llu", &n) != EOF) {

    ans_num = 0;

    dfs(0, 1, 1, 60);

    printf("%lld\n", ans);

    printf("%llu\n", ans);

  }

  return 0;

}