poj 2739 Sum of Consecutive Prime Numbers（尺取，暴力）

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本文探讨了如何找出一个正整数可以表示为一个或多个连续质数之和的方法及其实现。通过两种不同的算法实现——滑动窗口方法和暴力搜索方法，解决了日本2005年竞赛中的一道题目。

Sum of Consecutive Prime Numbers

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 23545		Accepted: 12862

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

Sample Output

Source

Japan 2005

因为数据很弱，所以暴力也可以过。

尺取的代码：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 1e4+5;
int a[maxn], prime[maxn], n, k;
int main(void)
{
    for(int i = 2; i < maxn; i++)
        if(!a[i])
        {
            prime[k++] = i;
            for(int j = 2; j*i < maxn; j++)
                a[i*j] = 1;
        }

    while(cin >> n, n)
    {
        int s = 0, e = 0, sum = 0, ans = 0;
        while(1)
        {
            while(e < k && sum < n)
                sum += prime[e++];
            if(sum < n) break;
            if(sum == n) ans++;
            sum -= prime[s++];
        }
        printf("%d\n", ans);
    }

    return 0;
}

暴力的代码：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 1e4+5;
int a[maxn], prime[maxn], k;
int main(void)
{
    for(int i = 2; i < maxn; i++)
        if(!a[i])
        {
            prime[k++] = i;
            for(int j = 2; j*i < maxn; j++)
                a[i*j] = 1;
        }
    int n, ans, sum;
    while(cin >> n, n)
    {
        ans = 0;
        for(int i = 0; i < k; i++)
        {
            sum = 0;
            for(int j = i; j < k; j++)
            {
                sum += prime[j];
                if(sum == n) ans++;
                if(sum > n) break;
            }
        }
        printf("%d\n", ans);
    }

    return 0;
}