POJ 2739 Sum of Consecutive Prime Numbers（素数序列和，尺取法）

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本文介绍了一道经典的算法题目，即求解一个正整数可以由多少种连续素数序列相加得到。通过筛选法预处理素数，并采用尺取法进行高效求解。

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

Sum of Consecutive Prime Numbers

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 23325		Accepted: 12749

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

题意：

给定一个数字，询问有多少个连续素数序列之和等于该值。

例如：

53 有两种：53=5 + 7 + 11 + 13 + 17；53=53；

但是20=7 + 13 和20=3 + 5 + 5 + 7因为素数序列不连续，所以不满足。

思路：

素数打表后利用尺取法求解。

AC Code：

#include<stdio.h>
#include<string.h>
const int MYDD=1e4+1103;

int primenum[MYDD];
bool isprime[MYDD];
int GetPrime() {//筛选法判断素数
	int num=0;
	memset(isprime,true,sizeof(isprime));
	isprime[0]=isprime[1]=false;
	for(int j=2; j<=MYDD; j++) {
		if(isprime[j]) {
			primenum[num]=j;
			for(int i=j*2; i<=MYDD; i=i+j)
				isprime[i]=false;
			num++;
		}
	}
	return num;//返回 10000 之内的素数个数
}


int main() {
	int dd=GetPrime();
	while(1) {
		int n;
		scanf("%d",&n);
		if(n==0)	break;
//		printf("%d*****\n",dd);
		int ans=0;
		int sum,l,r;
		l=r=sum=0;
		/*尺取法*/
		while(1) {
			while(r<dd&&sum<n)	sum+=primenum[r++];
			if(sum<n)		break;
			else if(sum==n)	ans++;
			sum-=primenum[l++];
		}
		printf("%d\n",ans);
	}
	return 0;
}