poj 2992 Divisors

　　筛素数，分解质因数。
代码如下：
 1 #include <cstdio>
 2 #include <iostream>
 3 #include <algorithm>
 4 #include <cstring>
 5 
 6 using namespace std;
 7 
 8 typedef long long LL;
 9 const int N = 444;
10 int prm[N >> 2], pn, pid[N], pv[N];
11 bool np[N];
12 
13 void getPrm() {
14     np[0] = np[1] = true;
15     prm[pn = 0] = 2, pid[2] = pn++;
16     for (int i = 3, tmp; i < N; i++, i++) {
17         if (!np[i]) prm[pn] = i, pid[i] = pn++;
18         for (int j = 0; j < pn; j++) {
19             tmp = prm[j] * i;
20             if (tmp >= N) break;
21             np[tmp] = true;
22             if (i % prm[j] == 0) break;
23         }
24     }
25     pv[0] = pv[1] = 1, pv[2] = 2;
26     for (int i = 2; i < N; i++, i++) pv[i] = 2;
27     for (int i = 3; i < N; i++, i++) {
28         if (!pv[i]) {
29             for (int j = i; j < N; j += i) pv[j] = i;
30         }
31     }
32 //    cout << pn << endl;
33 //    for (int i = 0; i < 10; i++) cout << prm[i] << endl;
34 //    for (int i = 0; i < 10; i++) cout << pv[i] << endl;
35 }
36 
37 const int M = 86;
38 LL dvs[N][N];
39 int tmp[M];
40 
41 void PRE() {
42     getPrm();
43     dvs[0][0] = dvs[1][0] = dvs[1][1] = 1;
44     for (int i = 2; i < N; i++) {
45         for (int j = 0; j < M; j++) tmp[j] = 0;
46         dvs[i][0] = 1;
47         for (int j = 1; j <= i; j++) {
48             int t = i - j + 1;
49             while (t > 1) {
50                 tmp[pid[pv[t]]]++;
51                 t /= pv[t];
52             }
53             t = j;
54             while (t > 1) {
55                 tmp[pid[pv[t]]]--;
56                 t /= pv[t];
57             }
58             dvs[i][j] = 1;
59             for (int k = 0; k < M; k++) dvs[i][j] *= tmp[k] + 1;
60 //            if (i == 412 && j == 205) { for (int i = 0; i < M; i++) cout << tmp[i] << ' '; cout << endl; cout << dvs[i][j] << endl;}
61         }
62     }
63 //    puts("ok~");
64 }
65 
66 int main() {
67     int n, m;
68     PRE();
69 //    cout << dvs[10][4] << endl;
70     while (~scanf("%d%d", &n, &m)) printf("%lld\n", dvs[n][m]);
71     return 0;
72 }
View Code
——written by Lyon
posted @ 2013-07-09 13:45 LyonLys 阅读(163) 评论(0) 收藏举报
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