【POJ2739】Sum of Consecutive Prime Numbers

简单的素数打表，然后枚举。开始没注意n读到0结束，TLE了回。。下次再认真点。A过后讨论里面有个暴力打表过的，给跪了！

 

 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 }