Sum of Consecutive Prime Numbers（前缀和，素数筛）

UVA1210

筛出素数，处理成前缀和， $O(n^2)$枚举每一种和的可能。

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <cstdio>
#include <cstring>
using namespace std;

const int MAXN = 10000;
bool IsPrime[MAXN + 1];
vector<int> PTable;
int sum[1230];

void InitPTable() {
	for (int i = 2; i <= MAXN; ++i) {
		if (IsPrime[i] == false) {
			PTable.push_back(i);
			for (int j = 2 * i; j <= MAXN; j += i) {
				IsPrime[j] = true;
			}
		}
	}
}

void InitPresum() {
	sum[1] = PTable[0];
	for (int i = 2; i <= 1229; ++i) {
		sum[i] = sum[i - 1] + PTable[i - 1];
	}
}

int main() {
	InitPTable();
	InitPresum();
	int n;
	while (cin >> n && n) {
		int Ans = 0;
		for (int i =1; i < 1229; ++i) {
			for (int j = 0; j < i; ++j) {
				Ans += (sum[i] - sum[j] == n);
			}
		}
		printf("%d\n", Ans);
	}
	return 0;

}