HDU-3189-Just Do It分解质因数

Posted 2020-07-07 宣之于口

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3189-Just Do It

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Problem Description
Now we define a function F(x), which means the factors of x. In particular, F(1) = 1,since 1 only has 1 factor 1, F(9) = 3, since 9 has 3 factors 1, 3, 9. Now give you an integer k, please find out the minimum number x that makes F(x) = k.

Input
The first line contains an integer t means the number of test cases.
The follows t lines, each line contains an integer k. (0 < k <= 100).

Output
For each case, output the minimum number x in one line. If x is larger than 1000, just output -1.

Sample Input
4
4
2
5
92

Sample Output
6
2
16
-1

题目连接：HDU-3189

题目大意：求约数个数为k的最小数。

题目思路：先求出每个数字的质因数，然后dfs求出所有的约数。

以下是代码：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <queue>
#include<iomanip>
//#include<bits\stdc++.h>
#define ll unsigned long long
using namespace std;
#define MAXN 1010
#define ANS_SIZE 505
ll f[ANS_SIZE],nf[ANS_SIZE];  //f存放质因数，nf存放对应质因数的个数
ll plist[MAXN], pcount=0;
bool isPrime[MAXN+1]; 
void initprime()//素数且不说，所有合数都能分解成任意素数之积
{
    int i,j;
    pcount = 0;
    for(i = 2; i<MAXN; i++)
    {
        if(!isPrime[i]) plist[pcount++] = i;//打下素数表
        for(int j = 0; j<pcount && i*plist[j]<MAXN; j++)
        {
            isPrime[i*plist[j]] = true;//所有非素数排除
            if(i%plist[j]==0) break;
        }
    }
}

int prime_factor(ll n) {
    int cnt = 0;
    int n2 = (int)sqrt((double)n);
    for(int i = 0; n > 1 && plist[i] <= n2 && i < pcount; ++i)
    {   
        if (n % plist[i] == 0) {            
            for (nf[cnt] = 0; n % plist[i] == 0; ++nf[cnt], n /= plist[i]);
            f[cnt++] = plist[i];
        }
    }
    if (n > 1) nf[cnt] = 1, f[cnt++] = n;
    return cnt;   //返回不同质因数的个数
}

vector<ll> yue[1005];
void dfs( int x, int t, ll ss ,int numpos){
    if( x==t ) return;
    dfs( x+1, t, ss,numpos);
    for( int i=0 ; i<nf[x] ; i++ ){
        ss *= f[x];
        yue[numpos].push_back(ss);
        dfs( x+1, t, ss,numpos);
    }
}
int ans[1005];
int main(){
    ll m;
    initprime();
    for (int i = 1; i < 1005; i++)
    {
        memset(f,0,sizeof(f));
        memset(nf,0,sizeof(nf));
        int ret = prime_factor(i);
        yue[i].push_back (1);
        dfs(0,ret,1,i);
        ans[i] = yue[i].size();
    }
    cin >> m;
    while(m--)
    {
        int k;
        cin >> k;
        int flag = 0;
        for (int i = 1; i < 1005; i++)
        {
            if (ans[i] == k)
            {
                flag = 1;
                cout << i << endl;
                break;
            }
        }
        if (!flag) cout << -1 << endl;
    }
    return 0;
}