hdu 2421 Deciphering Password-优快云博客

本文分享了解决HDU 2421问题的经验，通过质因数分解算法找到所有质因数，并利用埃拉托斯特尼筛法进行优化。介绍了关键步骤及调试过程。

http://acm.hdu.edu.cn/showproblem.php?pid=2421

　　因为没初始化质因数数组f，所以wa了好几遍....- - 还是初始化的问题，看来以后要注意点！

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 1 #include <cstdio>
 2 #include <cstring>
 3 #include <cstdlib>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <iostream>
 7 
 8 #define debug 0
 9 
10 using namespace std;
11 
12 typedef __int64 ll;
13 const int mod = 10007;
14 const int maxn = 1005;
15 
16 bool np[maxn];
17 int pn, pr[maxn >> 2];
18 
19 void gp(){
20     memset(np, 0, sizeof(np));
21     np[0] = np[1] = true;
22     pn = 0;
23     for (int i = 2; i < maxn; i++){
24         if (!np[i]) pr[pn++] = i;
25         for (int j = 0; j < pn && pr[j] * i < maxn; j++){
26             np[pr[j] * i] = true;
27             if (i % pr[j] == 0) break;
28         }
29     }
30     #if debug
31     printf("pn %d\n", pn);
32     #endif
33 }
34 
35 void fac(int a, int *f, int &cnt){
36     int i = 0;
37 
38     cnt = 0;
39     while (pr[i] * pr[i] <= a && i < pn){
40         while (a % pr[i] == 0){
41             f[cnt++] = pr[i];
42             a /= pr[i];
43         }
44         i++;
45     }
46     if (a != 1) f[cnt++] = a;
47 }
48 
49 void deal(int a, int b, int c){
50     int f[50], tmp;
51     int ans = 1;
52     int cnt = 0;
53     int n = 0;
54 
55     memset(f, 0, sizeof(f));
56     b %= mod;
57     fac(a, f, n);
58     sort(f, f + n);
59     #if debug
60     printf("n  %d\n\n", n);
61     for (int i = 0; i < n; i++){
62         printf("%d\n", f[i]);
63     }
64     f[n] = 0;
65     #endif
66     for (int i = 0; i < n; i++){
67         cnt++;
68         if (f[i] != f[i + 1]) {
69             #if debug
70             printf("cnt %d   b %d\n", cnt, b);
71             #endif
72             cnt *= b;
73             cnt %= mod;
74             tmp = (((cnt + 1) % mod) * (cnt + 2) / 2) % mod;
75             tmp = (tmp * tmp) % mod;
76             ans *= tmp;
77             ans %= mod;
78             cnt = 0;
79         }
80         if (!ans) break;
81     }
82     printf("Case %d: %d\n", c, ans);
83 }
84 
85 int main(){
86     int i = 1;
87     int a, b;
88 
89     gp();
90     while (~scanf("%d%d", &a, &b)){
91         deal(a, b, i);
92         i++;
93     }
94 
95     return 0;
96 }