POJ_2739_Sum_of_Consecutive_Prime_Numbers_(尺取法+素数表)

描述

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http://poj.org/problem?id=2739

多次询问,对于一个给定的n,求有多少组连续的素数,满足连续素数之和为n.

 

Sum of Consecutive Prime Numbers

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 22737		Accepted: 12432

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

Source

Japan 2005

分析

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尺取法.

打个素数表,对于每一个n,尺取之.

换了种尺取法的写法= =.

#include<cstdio>

const int maxn=10005;
int p,n;
int prime[maxn];
bool is_prime[maxn];

void get_prime()
{
    for(int i=1;i<maxn;i++) is_prime[i]=true;
    is_prime[0]=is_prime[1]=false;
    p=0;
    for(int i=2;i<maxn;i++)
    {
        if(is_prime[i])
        {
            prime[++p]=i;
            for(int j=2*i;j<maxn;j+=i) is_prime[j]=false;
        }
    }
}

void solve()
{
    int l=1,r=0,sum=0,ans=0;
    while(l<=p&&r<=p&&prime[l]<=n&&prime[r]<=n)
    {
        if(sum==n) { ans++; sum-=prime[l++]; }
        else if(sum>n) sum-=prime[l++];
        else sum+=prime[++r];
    }
    printf("%d\n",ans);
}

void init()
{
    get_prime();
    while(scanf("%d",&n)==1&&n!=0) solve();
}

int main()
{
    freopen("sum.in","r",stdin);
    freopen("sum.out","w",stdout);
    init();
    fclose(stdin);
    fclose(stdout);
    return 0;
}