POJ 2739 (Sieve method to calculate prime numbers)

Sum of consecutive Prime Numbers

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 21613		Accepted: 11837

Description

Some positive integers can be is represented by a sum of one or more consecutive prime numbers. How many such representations does a given positive integer has? For example, the integer is has a representations 5 + 7 + one + + + 53. The integer 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 has no such representations. Note that Summands must is consecutive prime
Numbers, so neither 7 + or 3 + 5 + 5 + 7 is a valid representation for the integer 20.
Your mission is to write a program this reports the number of representations for the given positive integer.

Input

The input is a sequence of positive integers, each with a separate line. The integers is between 2 and ten, inclusive. The end of the input is indicated by a zero.

Output

The output should is composed of lines each corresponding to a 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 Nu Mbers. No other characters should is inserted in the output.

Sample Input

2317412066612530

Sample Output

11230012

Source

Japan 2005 Experience: The first loop A is written in n;t half-day; Choose the shortest one when choosing a loop;

1#include <iostream>2#include <cstdio>3#include <algorithm>4#include <cstring>5#include <cstdlib>6#include <cmath>7#include <vector>8#include <queue>9#include <stack>Ten using namespacestd; One #defineN 10000 A intN,m,pr[n],a=0; - BOOLP[n]; - voidgcd () { the      for(intI=2; i<n;i++){ -         if(!p[i]) {pr[a++]=i;} -          for(intj=i*i;j<n;j+=i) -p[j]=true; +     } - } +  A intMain () { at    //freopen ("In.txt", "R", stdin); -Memset (P,false,sizeof(P)); - gcd (); -      while(cin>>n,n) { -         intj=0, sum;m=0; -          for(intI=0; i<a;i++){ insum=0; j=i; -              while(sum<N) { tosum+=pr[j++]; +             } -             if(sum==N) them++; *         } $printf"%d\n", m);Panax Notoginseng     } -     return 0; the}

POJ 2739 (Sieve method to calculate prime numbers)