【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！

 #include <iostream>
 #include <cstdlib>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <cctype>
 #include <complex.h>
 #include <cmath>
 using namespace std;

 const int N = ;

 bool is[N]; int prm[N];
 int getprm (int n) {
     int i, j, k = ;
     int s, e = (int)(sqrt(0.0 + n) + );
     memset (is, , sizeof (is));
     prm[k ++] = ; is[] = is[] = ;
     for (i = ; i < n; i += ) is[i] = ;
     for (i = ; i < e; i += )
         if (is[i]){
             prm[k ++] = i;
             for (s = i * , j = i * i; j < n; j += s) {
                 is[j] = ;
         }
     }
     for (; i < n; i += ) {
         if (is[i]) prm[k ++] = i;
     }
     return k;
 }

 int main () {
     getprm();
     /*for (int i = 0 ; i < 100; ++ i) {
         cout << i << ":  ";
         cout << prm[i] << endl;
     }*/
     // 1229个素数 0-1w
     ios::sync_with_stdio(false);
     cin.tie();
     int n;
     while (~scanf ("%d", &n)) {
         if (n == ) break;
         int sum, cnt = ;
         for (int i =  ; prm[i] <= n; ++ i) {
             sum = ;
             for (int j = i; prm[j] <= n; ++ j) {
                 sum += prm[j];
                 if (sum < n) {
                     continue;
                 } else if (sum == n) {
                     cnt ++;
                     break;
                 } else {
                     break;
                 }
             }
         }
         printf ("%d\n", cnt);
     }
     return ;
 }