ACM：POJ 2739 Sum of Consecutive Prime Numbers-素数打表-尺取法

POJ 2739 Sum of Consecutive Prime Numbers

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu

 

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：

求连续素数和.

有多少种方法可以选取连续的素数，使这些数的和正好为n

思路：

1到10000的素数表打出来，然后直接尺取就可以了，很简单的一道题目。
/*/

#include"map"
#include"cmath"
#include"string"
#include"cstdio"
#include"vector"
#include"cstring"
#include"iostream"
#include"algorithm"
using namespace std;
typedef long long LL;
const int MX=1000005;
#define memset(x,y) memset(x,y,sizeof(x))
#define FK(x) cout<<"【"<<x<<"】"<<endl

int vis[MX];
int prim[MX];
int ans[MX];
int main() {
	int n,len=0,sum=0;
	for(int i=2; i<=100; i++)
		if(!vis[i])
			for(int j=2; j<=10000; j++)
				vis[j*i]=1;
	for(int i=2; i<=10000; i++)
		if(vis[i]==0)    {
			len++;
			prim[len]=i;
		}
	for(int i=1; i<=len; i++)   {
		sum=0;
		int j=i;
		while(sum<=10000 && j<=1229) {
			sum+=prim[j];
			if(sum>10000)
				break;
			ans[sum]++;
			j++;
		}
	}
	while(~scanf("%d",&n)) {
		if(!n)break;
		printf("%d
",ans[n]);
	}
	return 0;
}