Codeforces Round #696 (Div. 2) B. Different Divisors

Codeforces Round #696 (Div. 2) B. Different Divisors

原题链接)

题目分析

找规律，最后会发现是间隔大于等于d的两个质数的积。可以先打表，然后再用二分查找找到对应的下标，输出答案即可。

AC代码

#include<bits/stdc++.h>
using namespace std;
#define io ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)

const int N = 1e5 + 1000;
bool isnp[N];
int k ;
int prime[N]; 
void init(){
    for (int i = 2; i <= N; i++){
        if (!isnp[i])  prime[k++] = i;
        for (int j = 0; j < N; j++){
            if (prime[j] * i > N)  break;
            isnp[prime[j] * i] = 1;
            if (i % prime[j] == 0)  break;
        }
    }
}

int t, n, m;
int main(){
    io;
    init();
    cin >> t;
    while(t--){
    	cin >> n;
    	int a = lower_bound(prime, prime + k, n + 1) - prime;
    	int b = lower_bound(prime + a, prime + k, n + prime[a]) - prime;
    	long long ans = prime[a] * prime[b];
    	cout << ans << endl;
    }

    return 0;
}