poj 2379 Sum of Consecutive Prime Numbers

                                                                                                         Sum of Consecutive Prime Numbers

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 24498		Accepted: 13326

Description

Some positive integers can be represented by a sum of one or more consecutive prime numbers. How many such representations does a given positive integer have? For example, the integer 53 has two representations 5 + 7 + 11 + 13 + 17 and 53. The integer 41 has three representations 2+3+5+7+11+13, 11+13+17, and 41. The integer 3 has only one representation, which is 3. The integer 20 has no such representations. Note that summands must be consecutive prime
numbers, so neither 7 + 13 nor 3 + 5 + 5 + 7 is a valid representation for the integer 20.
Your mission is to write a program that reports the number of representations for the given positive integer.

Input

The input is a sequence of positive integers each in a separate line. The integers are between 2 and 10 000, inclusive. The end of the input is indicated by a zero.

Output

The output should be composed of lines each corresponding to an input line except the last zero. An output line includes the number of representations for the input integer as the sum of one or more consecutive prime numbers. No other characters should be inserted in the output.

Sample Input

2
3
17
41
20
666
12
53
0

Sample Output

1
1
2
3
0
0
1
2

题意：某些数可以是连续质数的和，那么给定数a,求a的分解有几种可能。
思路：先可以用埃氏筛选找出连续的质数，之后尺取法找可能的情况，值得注意的是两组可能的连续质数序列存在元素重叠的情况，所以每找到一组，尺取法的头部和尾部统一更新为上一种情况的尾部再加1，继续查找下一种情况。。。
AC代码：

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<algorithm>
using namespace std;
const int N_MAX = 10000 + 4;
int prime[N_MAX],res;
bool is_prime[N_MAX+1],what;
int sieve(int n) {
    int p = 0;
    for (int i = 0; i <= n; i++)is_prime[i] = true;
    is_prime[0] = is_prime[1] = false;
    for (int i = 2; i <= n; i++) {
        if (is_prime[i]) {
            prime[p++] = i;
            for (int j = i*i; j <= n; j += i)is_prime[j] = false;
        }
    }
    return  p;
}
int main() {
    int n=sieve(N_MAX),a;
    while (scanf("%d",&a)&&a) {
        res = 0,what = 0;
        int l= 0, r= 0, sum = 0;
        int bound=upper_bound(prime, prime + n, a)-prime;
        while (1) {////////
            for (;;) {////
                while (r < bound&&sum < a) {
                    sum += prime[r++];
                }
                if (sum == a) {
                    res++;
                    what = 1;
                    break;
                }
                if (sum < a) {//找不到了，终止搜索；
                    what = 0;
                    break;
                }
                sum -= prime[l++];
            }////
            if (what) { l++; r = l; sum = 0; }
            else break;
        }////////
        printf("%d
", res);
    }
    return 0;
}