POJ 2739 尺取法

题意

传送门 POJ 2739

题解

埃氏筛法筛素数，然后尺取法推进头尾搜索，统计满足条件的可能求和值个数即可。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
#define min(a,b)    (((a) < (b)) ? (a) : (b))
#define max(a,b)    (((a) > (b)) ? (a) : (b))
#define abs(x)    ((x) < 0 ? -(x) : (x))
#define INF 0x3f3f3f3f
#define delta 0.85
#define eps 1e-5
#define PI 3.14159265358979323846
#define MAX_N 10005
using namespace std;
int N, cnt;
int rec[MAX_N];
bool is_prime[MAX_N];
int prime[MAX_N / 2];

void sieve(int n){
	int limit = sqrt((double)n);
	for(int i = 2; i <= n; i++) is_prime[i] = 1;
	for(int i = 2; i <= n; i++){
		if(is_prime[i]){
			prime[cnt++] = i;
			if(i <= limit){
				for(int j = i * i; j <= n; j += i) is_prime[j] = 0;
			}
		}
	}
}

int calc(int n){
	int limit = upper_bound(prime, prime + cnt, n) - prime;
	int s = 0, t = 0, sum = 0, res = 0;
	for(;;){
		while(t < limit && sum < n){
			sum += prime[t++];
		}
		if(sum == n) ++res;
		else if(sum < n) break;
		sum -= prime[s++];
	}
	return rec[n] = res;
}

int main(){
	sieve(MAX_N);
	memset(rec, -1, sizeof(rec));
	while(~scanf("%d", &N) && N){
		printf("%d\n", rec[N] == -1 ? calc(N) : rec[N]);
	}
	return 0;
}