【POJ2739】Sum of Consecutive Prime Numbers

 1 #include <iostream>
 2 #include <cstdlib>
 3 #include <cstdio>
 4 #include <cstring>
 5 #include <algorithm>
 6 #include <cctype>
 7 #include <complex.h>
 8 #include <cmath>
 9 using namespace std;
10 
11 const int N = 10005;
12 
13 bool is[N]; int prm[N];
14 int getprm (int n) {
15     int i, j, k = 0;
16     int s, e = (int)(sqrt(0.0 + n) + 1);
17     memset (is, 1, sizeof (is));
18     prm[k ++] = 2; is[0] = is[1] = 0;
19     for (i = 4; i < n; i += 2) is[i] = 0;
20     for (i = 3; i < e; i += 2)
21         if (is[i]){
22             prm[k ++] = i;
23             for (s = i * 2, j = i * i; j < n; j += s) {
24                 is[j] = 0;
25         }
26     }
27     for (; i < n; i += 2) {
28         if (is[i]) prm[k ++] = i;
29     }
30     return k;
31 }
32 
33 int main () {
34     getprm(10000);
35     /*for (int i = 0 ; i < 100; ++ i) {
36         cout << i << ":  ";
37         cout << prm[i] << endl;
38     }*/
39     // 1229个素数 0-1w
40     ios::sync_with_stdio(false);
41     cin.tie(0);
42     int n;
43     while (~scanf ("%d", &n)) {
44         if (n == 0) break;
45         int sum, cnt = 0;
46         for (int i = 0 ; prm[i] <= n; ++ i) {
47             sum = 0;
48             for (int j = i; prm[j] <= n; ++ j) {
49                 sum += prm[j];
50                 if (sum < n) {
51                     continue;
52                 } else if (sum == n) {
53                     cnt ++;
54                     break;
55                 } else {
56                     break;
57                 }
58             }
59         }
60         printf ("%d\n", cnt);
61     }
62     return 0;
63 }