Codeforces 762A k-th divisor(数论)

题目链接：k-th divisor

求出N的第K大因子，满足N <= 10^15，K <= 10^9

直接暴力……

 

 1 #include <bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 #define rep(i,a,b)              for(int i(a); i <= (b); ++i)
 6 #define LL              long long
 7 
 8 LL n, k, h, ans;
 9 int num;
10 
11 int main(){
12 
13     scanf("%lld%lld", &n, &k);
14     h = (LL)sqrt(n + 0.5);
15     if (h * h == n){
16         rep(i, 1, h - 1) if (n % i == 0) num += 2;
17         ++num;
18     }
19     else{ rep(i, 1, h) if (n % i == 0) num += 2; }
20 
21     if (num < k){ puts("-1"); return 0; }
22 
23     if (h * h == n){
24         int cnt = 0;
25         if (k <= num / 2){
26             rep(i, 1, h) if (n % i == 0){
27                 ++cnt;
28                 if (cnt == k){
29                     ans = i;
30                     break;
31                 }
32             }
33         }
34 
35         else if (k == num / 2 + 1){
36             printf("%lld\n", h);
37             return 0;
38         }
39 
40         else{
41             int m = k - num / 2 - 1;
42             int ret = num / 2 - m + 1;
43             rep(i, 1, h) if (n % i == 0){
44                 ++cnt;
45                 if (cnt == ret){
46                     ans = n / i;
47                     break;
48                 }
49             }
50         }
51     }
52 
53 
54 
55     else
56     {
57         int cnt = 0;
58         if (k <= num / 2){
59             rep(i, 1, h) if (n % i == 0){
60                 ++cnt;
61                 if (cnt == k){
62                     ans = i;
63                     break;
64                 }
65             }
66         }
67 
68         else{
69             int m = k - num / 2, cnt = 0;
70             int ret = num / 2 - m + 1;
71             rep(i, 1, h) if (n % i == 0){
72                 ++cnt;
73                 if (cnt == ret){
74                     ans = n / i;
75                     break;
76                 }
77             }
78         }
79     }
80 
81     printf("%lld\n", ans);
82     return 0;
83 
84 }

 

posted @ 2017-01-29 12:55  cxhscst2  阅读(374)  评论(0)  编辑  收藏  举报