POJ：2739-Sum of Consecutive Prime Numbers(尺取)

Sum of Consecutive Prime Numbers

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 27853 Accepted: 14968

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

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解题心得：

1. 题意就是一个数可以由多个连续的素数相加得到，现在给你一个素数ｎ，问你有几种由连续素数相加的方案。
2. 先素数筛选，把所有的素数放在一个数组里面，然后用尺取法在里面跑就行了。

---

#include <algorithm>
#include <stdio.h>
#include <string>
#include <vector>
using namespace std;
const int maxn = 1e4+100;
vector <int> ve;
int n;
bool vis[maxn];

void pre_deal() {
    for(int i=2;i<=maxn;i++) {
        if(vis[i])
            continue;
        ve.push_back(i);
        for(int j=i*2;j<=maxn;j+=i) {
            vis[j] = true;
        }
    }
}

int cal_num() {
    int l = 0,r = 0,sum = 0,cnt = 0;
    while(r < ve.size()) {
        while(sum < n && r < ve.size()) {
            sum += ve[r];
            r++;
        }
        if(sum < n)
            break;
        if(sum == n)
            cnt++;
        sum -= ve[l];
        l++;
    }
    return cnt;
}

int main() {
    pre_deal();
    while(scanf("%d",&n) && n) {
        printf("%d\n",cal_num());
    }
    return 0;
}