B - Sum of Consecutive Prime Numbers

最新推荐文章于 2025-03-17 16:51:00 发布

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最新推荐文章于 2025-03-17 16:51:00 发布

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分类专栏： oj 文章标签：算法 cpp 尺取法

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这篇博客讨论了一种寻找正整数的不同表示方法，即如何将其表示为一个或多个连续素数之和。文章介绍了尺取法来解决这个问题，并提供了C++代码实现。通过输入正整数，程序计算并输出该数有多少种这样的连续素数表示形式。例如，53有两种表示，而20没有。博客内容涉及到了素数判断、区间移动和计数策略等概念。

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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

题解：
尺取法
使用尺取法时要注意，区间如何移动，我们要选取的是什么样的区间

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#define INF 0x3f3f3f3f
using namespace std;
int prime[10005];
bool IsPrime(int n){
    int sum=0;
    for(int i=2;i<n;i++){
        if(n%i==0){
            sum++;
            break;
        }
    }
    if(sum==0){
        return true;
    }
    else{
        return false;
    }

}
int main(){
    std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    int cnt=0;
    for(int i=2;i<=10000;i++){
        if(IsPrime(i)){
            prime[cnt++]=i;
        }
    }
    prime[cnt]=INF;
    //cnt最后的结果存的是素数的个数
    //cout<<prime[cnt-1]<<endl;
    //cout<<prime[cnt-2]<<endl;
    int n;
    while(cin>>n){
        int sum=0;//和
        int st=0,en=0;//左右端点
        int ans=0;//个数
        if(n==0){
            break;
        }
        int end=cnt;
        //end为区间右端点
        while(1){
            while(en<=end&&sum<n){
                sum+=prime[en++];
            }
            if(sum<n){
                break;
            }
            else if(sum==n){
                ans++;
            }
            sum-=prime[st++];
        }
        cout<<ans<<endl;

    }
    return 0;
}