POJ2739 Sum of Consecutive Prime Numbers（AC代码 + 详解）

• Sum of Consecutive Prime Numbers

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

题目的意思是一个正整数可以由一个或者多个连续素数表示，给出一个正整数它有多少种表示呢？例如53可以由5+7+11+13+17和53两种方法表示，41有2+3+5+7+11+13和11+13+17和41表示，20没有表示的方法……当输入0时表示终止输入（大概就是这个意思，只能是由连续素数组成，不能重复出现同一素数）数据范围：2~10000
这个题先进行预处理再从左到右扫描素数数组就可AC
下面是AC代码（附注解）

/*================================================================
# Copyright (C) 2020 Defepe. All rights reserved.
# File Name: poj2739.cpp
# Author: Defepe
# Created Time: 2020/7/3 16:23:31
================================================================*/
#include 

using namespace std;

int prime[10005] = {0}, vis[10005] = {0}, cnt = 0;
void func() {
	for(int i = 2; i <= 10000; i++) {
		if(!vis[i]) prime[cnt++] = i;
		for(int j = 0; j < cnt && prime[j] * i <= 10000; j++) {
			vis[prime[j] * i] = 1;
			if(i % prime[j] == 0) break;
		}
	}
}
int main() {
	func();   //先进行预处理，计算出2~10000内的素数表，这里用的是欧拉素数筛
	int n, ans = 0;  //ans是方法数，n为输入的整数
	while(scanf("%d", &n) && n) {  //当n为0时退出循环
		for(int i = 0; prime[i] <= n; i++)  //这两层循环相当于做了一个从左到右的扫描，以i为左端点，j为右端点进行扫描，如果相等ans++，如果大于则左端点向右移动，如果小于则右端点向右移动
		{
			int sum = 0;
			for(int j = i; j < cnt; j++)
			{
				sum += prime[j];
				if(sum == n) ans++;
				else if(sum > n) break;
			}
		}
		printf("%d\n", ans);  //输出这个数的答案的同时也要将ans赋0，进行下一个数的计算
		ans = 0;
	}
	return 0;
}

如果还有什么不懂的地方欢迎在评论区留言，有错误的地方也希望大家能够指出，谢谢