False Ordering LightOJ

We define b is a Divisor of a number a if a is divisible by b. So, the divisors of 12 are 1, 2, 3, 4, 6, 12. So, 12 has 6 divisors.

Now you have to order all the integers from 1 to 1000. x will come before y if

1)                  number of divisors of x is less than number of divisors of y

2)                  number of divisors of x is equal to number of divisors of y and x > y.

Input

Input starts with an integer T (≤ 1005), denoting the number of test cases.

Each case contains an integer n (1 ≤ n ≤ 1000).

Output

For each case, print the case number and the n^th number after ordering.

Sample Input

5

1

2

3

4

1000

Sample Output

Case 1: 1

Case 2: 997

Case 3: 991

Case 4: 983

Case 5: 840

直接分解质因子 暴力就好了。。。

#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#define rap(a, n) for(int i=1; i<=n; i++)
#define MOD 2018
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _  ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
const int maxn = 10010, INF = 0x7fffffff;
int primes[maxn], vis[maxn], sum[maxn];
int ans;

struct node
{
    int id, xi;
}Node[maxn];


void init()
{
    mem(vis, 0);
    for(int i=2; i<maxn; i++)
    {
        if(vis[i]) continue;
        primes[ans++] = i;
        for(LL j = (LL)i*i; j<maxn; j+=i)
            vis[j] = 1;
    }
}

bool cmp(node a, node b)
{
    if(a.xi == b.xi)
        return a.id > b.id;
    return a.xi < b.xi;
}

int main()
{
    ans = 0;
    init();
    rap(1, 1000)
    {
        Node[i].id = i;
        Node[i].xi = 1;
        int temp = i;
        for(int j=0; j<ans && primes[j] * primes[j] <= temp; j++)
        {
            int cnt2 = 0;
            while(temp % primes[j] == 0)
            {
                temp /= primes[j];
                cnt2++;
            }
            if(cnt2 > 0)
                Node[i].xi *= (cnt2+1);
        }
        if(temp > 1)
            Node[i].xi *= 2;
    }
    sort(Node+1, Node+1+1000, cmp);
    int T, kase = 0;
    cin>> T;
    while(T--)
    {
        int n;
        cin>> n;
        printf("Case %d: %d
",++kase, Node[n].id);
    }

    return 0;
}