codeforce 1474B Different Divisors 模拟 素数筛 C

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codeforce 1474B Different Divisors 模拟 素数筛 C

B. Different Divisors

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Positive integer xx is called divisor of positive integer yy, if yy is divisible by xx without remainder. For example, 11 is a divisor of 77 and 33 is not divisor of 88.

We gave you an integer dd and asked you to find the smallest positive integer aa, such that

• aa has at least 44 divisors;
• difference between any two divisors of aa is at least dd.

Input

The first line contains a single integer tt (1≤t≤30001≤t≤3000) — the number of test cases.

The first line of each test case contains a single integer dd (1≤d≤100001≤d≤10000).

Output

For each test case print one integer aa — the answer for this test case.

Example

input

Copy

2
1
2

output

Copy

6
15

Note

In the first test case, integer 66 have following divisors: [1,2,3,6][1,2,3,6]. There are 44 of them and the difference between any two of them is at least 11. There is no smaller integer with at least 44 divisors.

In the second test case, integer 1515 have following divisors: [1,3,5,15][1,3,5,15]. There are 44 of them and the difference between any two of them is at least 22.

The answer 1212 is INVALID because divisors are [1,2,3,4,6,12][1,2,3,4,6,12]. And the difference between, for example, divisors 22 and 33 is less than d=2d=2.

 

 

 

分析

要求两个因子之间至少差d，至少四个，最好是素数

那数据1e4。也不大，打表素数，挨个寻找，找到两个就可以了。

 

代码

 https://codeforces.com/contest/1474/submission/105586801

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <math.h>
#include <string.h>
#include <algorithm>
#include <iostream>
#include <string>
#include <sstream>
#include <iostream>
#include <time.h>
#include <queue>
#include <list>
#include <map>
#include <set>
#include <vector>
#include <stack>
#include <string.h>
#include <bitset>
#define sf scanf
#define pf printf
#define lf double
#define p123 printf("123\n");
#define pn printf("\n");
#define pk printf(" ");
#define p(n) printf("%d",n);
#define pln(n) printf("%d\n",n);
#define s(n) scanf("%d",&n);
#define ss(n) scanf("%s",n);
#define ps(n) printf("%s",n);
#define sld(n) scanf("%lld",&n);
#define pld(n) printf("%lld",n);
#define slf(n) scanf("%lf",&n);
#define plf(n) printf("%lf",n);
#define sc(n) scanf("%c",&n);
#define pc(n) printf("%c",n);
#define gc getchar();
#define ll long long
#define re(n,a) memset(n,a,sizeof(n));
#define len(a) strlen(a)
#define eps 1e-13
#define zero(x) (((x) > 0? (x):(-x)) < eps)
using namespace std;
int a[40000];
int main(){
    int t;
    s(t)
    int maxi = 30505;
    for(int i = 0; i <= maxi; i ++){
        a[i] = 1;
    }
    for(int i = 2; i <= maxi; i ++){
        if(a[i] == 1){
            for(int j = i*i; j <= maxi; j += i){
                a[j] = 0;
            }
        }
    }
    while(t --){
        int n;
        s(n)
        int x,y;
        for(int i = 1+n; i <= maxi; i ++){
            if(a[i] == 1){
                x = i;
                break;
            }
        }
        for(int i = x+n; i <= maxi; i ++){
            if(a[i] == 1){
                y = i;
                break;
            }
        }
        p(x*y) pn
    }
    return 0;
}