Sum of Consecutive Prime Numbers

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 14728 Accepted: 8197

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
解:
这道题每组测试数据都是要计算素数的,范围是2-10000,所以我们可以先构建素数表,然后再来计算每个测试数据可以由几组连续的素数构成。其中外循环是用来枚举所有可能的素数,内循环是用来将素数加到等于或者大于输入的测试数据为止。
#include <iostream>
#include <stdio.h>
using namespace std;

int prime[1229]; //这里的长度是后来计算出来的,为了省内存。
int count = 0;

//判断k是否为素数,模它前面的素数,如果得到0说明它不是素数,反之这是素数。
bool isPrime(int k) 
{
    for (int i = 0; i < count; i++) {
        if (k % prime[i] == 0) {
            return false;
        }
    }
    return true;
}

int main()
{
    //开始构建素数表
    for (int i = 2; i <= 10000; i++) {
        if (isPrime(i)) {
            prime[count] = i;
            count++;
        }
    }

    int input;
    while (cin>>input) {
        if (input == 0) {
            return 0;
        }
        int total = 0;
        for (int i = 0; prime[i] <= input; i++) {
            int sum = 0;
            for (int j = i; sum <= input; j++) {
                sum += prime[j];
                if (sum == input) {
                    total++;
                    break;
                }
            }
        }
        cout<<total<<endl;
    }
}
Result Memory Time Language Code Length
Accepted 252K 32MS C++ 672B

 

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