先打素数表,然后打表就OK了。水题一个。
POJ 2739
/*
MDK | 2739 | Accepted | 796K | 0MS | G++ | 2305B | 2011-11-03 21:45:41 |
*/
LL mat[MAXN];
bool is[MAXN];
int prm[MAXN];
int getprm(int n)
{
int i, j, k = 0;
int s, e = (int)(sqrt(0.0 + n) + 1);
memset(is, 1, sizeof(is));
prm[k++] = 2;
is[0] = is[1] = 0;
for (i = 4; i < n; i += 2) is[i] = 0;
for (i = 3; i < e; i += 2) if (is[i]) {
prm[k++] = i;
for (s = i * 2, j = i * i; j < n; j += s)
is[j] = 0;
// 因为j是奇数,所以+奇数i后是偶数,不必处理!
}
for ( ; i < n; i += 2) if (is[i]) prm[k++] = i;
return k; // 返回素数的个数
}
void init()
{
SET(mat,0);
int len = getprm(MAXN);
LL sum;
F(i,len) {
sum =0;
for(int k = i;k < len;k++) {
sum += prm[k];
if(sum > MAXN) break;
mat[sum]++;
}
}
}
int main ()
{
init();
int n;
while(~SCF(n) && n) {
printf("%d\n",mat[n]);
}
}