Sum of Consecutive Prime Numbers（素数打表+尺取）

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


题目意思：给你一个数问你这个数可以有多少种方案的连续素数来组成，输出方案数。

解题思路：素数打表与尺取的组合应用。

  https://www.cnblogs.com/wkfvawl/p/9092546.html

 

开始以为应该时间超不了限，暴力了一发，果然还真是出问题了

#include<stdio.h>
#include<string.h>
#define MAX 10010
long long s[MAX],isprime[MAX];
void prime()///素数打表
{
    long long i,k,j;
    k=1;
    memset(isprime,1,sizeof(isprime));///初始化都认为是素数
    isprime[0]=0;
    isprime[1]=0;///0和1不是素数
    for(i=2; i<=MAX; i++)
    {
        if(isprime[i])
        {
            s[k++]=i;///保存素数
        }
        for(j=i*2; j<=MAX; j+=i)
        {
            isprime[j]=0;///素数的倍数都不是素数
        }
    }
}
int main()
{
    int n,i,j,count,sum;
    prime();
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0)
        {
            break;
        }
        count=0;
        for(i=1; i<10010; i++)
        {
            sum=0;
            for(j=i; j<10010; j++)
            {
                sum=sum+s[j];
                if(sum==n)
                {
                    count++;
                    break;
                }
                else if(sum>n)
                {
                    break;
                }
            }
        }
        printf("%d\n",count);
    }
    return 0;
}

 

这是正确的尺取法的答案

 1 #include<stdio.h>
 2 #include<string.h>
 3 #define MAX 10010
 4 long long s[MAX],isprime[MAX];
 5 void prime()///素数打表
 6 {
 7     long long i,k,j;
 8     k=1;
 9     memset(isprime,1,sizeof(isprime));///初始化都认为是素数
10     isprime[0]=0;
11     isprime[1]=0;///0和1不是素数
12     for(i=2; i<=MAX; i++)
13     {
14         if(isprime[i])
15         {
16             s[k++]=i;///保存素数
17         }
18         for(j=i*2; j<=MAX; j+=i)
19         {
20             isprime[j]=0;///素数的倍数都不是素数
21         }
22     }
23 }
24 int main()
25 {
26     int n,i,j,count,sum;
27     prime();
28     while(scanf("%d",&n)!=EOF)
29     {
30         if(n==0)
31         {
32             break;
33         }
34         count=0;
35         sum=0;
36         j=1;
37         i=0;
38         while(1)
39         {
40             while(sum<n&&s[i+1]<=n)///s[i+1]<=n表示这个数是可以加的，即右端点还可以继续右移
41             {
42                 sum=sum+s[++i];
43             }///开始先找到一个满足条件的序列，之后扩大左端点
44             if(sum<n)
45             {
46                 break;
47             }
48             else if(sum>n)
49             {
50                 sum=sum-s[j];
51                 j++;
52             }///不断扩大右端点
53             else if(sum==n)///存在这样一个连续素数的集合
54             {
55                 count++;
56                 sum=sum-s[j];///继续扩大右端点
57                 j++;
58             }
59         }
60         printf("%d\n",count);
61     }
62     return 0;
63 }