CodeForces - 762A k-th divisor

3923 단어 onlinejudgeCodeForces
A. k-th divisor
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
You are given two integers n and k. Find k-th smallest divisor of n, or report that it doesn’t exist.
Divisor of n is any such natural number, that n can be divided by it without remainder.
Input
The first line contains two integers n and k (1 ≤ n ≤ 1015, 1 ≤ k ≤ 109).
Output
If n has less than k divisors, output -1.
Otherwise, output the k-th smallest divisor of n.
Examples
Input
4 2

Output
2

Input
5 3

Output
-1

Input
12 5

Output
6

Note
In the first example, number 4 has three divisors: 1, 2 and 4. The second one is 2.
In the second example, number 5 has only two divisors: 1 and 5. The third divisor doesn’t exist, so the answer is -1.
제목: 두 개의 숫자 n, k를 주고 첫 번째 숫자의 k번째 인수를 구한다. 만약 이 출력이 없다면 - 1蒟蒻의 블로거는 제목의 뜻을 잘못 보았다고 cf가 성공적으로 0이 되었는가
#include
#include
#define LL long long

int main()
{
    LL n, k;
    while(~scanf("%lld%lld",&n,&k)) {
        LL x = sqrt(n);
        bool flag = false;
        LL ans_1 = 0, ans = 0;
        for(LL i = 1;i <= x; i++) {
            if(n%i == 0) 
                ans++;
                if(ans == k) {
                printf("%lld
"
,i); flag = true; break; } } if(!flag){ if(x*x == n) { ans = ans*2 - 1; } else { ans = ans*2; } if(ans < k) { printf("-1
"
); flag = true; } } if(!flag) { ans = ans + 1 -k; for (LL i = 1;i <= x; i++) { if(n%i == 0) { ans_1++; if(ans_1 == ans) { printf("%lld
"
,(LL)n/i); break; } } } } } return 0; }

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