POJ 2739 Sum of Consecutive Prime Numbers(数论)

Sum of Consecutive Prime Numbers

http://poj.org/problem?id=2739

Time Limit:  1000MS

Memory Limit: 65536K

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

Source

Japan 2005

思路：两个for，可以暴力，若想优化，详见代码。

完整代码：

/*0ms,180KB*/

#include
const int maxn = 10001;

int prime[1229], c;
bool vis[maxn];

inline void create_prime()
{
	for (int i = 2; i < maxn; ++i)
		if (!vis[i])
		{
			prime[c++] = i;
			for (int j = i * i; j < maxn; j += i)
				vis[j] = true;
		}
}

int main()
{
	create_prime();
	//printf("##%d",c);
	int n, temp, i;
	while (scanf("%d", &n), n)
	{
		int count = 0;
		if (n > 3 && (n & 1 == 0))///大于4的偶数肯定不行~
		{
			puts("0");
			continue;
		}
		temp = n >> 1;
		for (i = 0; prime[i] <= temp; ++i)///算一半就行~
		{
			int sum = prime[i];
			for (int j = i + 1; sum < n; ++j)
				sum += prime[j];
			if (sum == n)
				count++;
		}
		for (; prime[i] < n; ++i)
			;
		if (prime[i] == n)
			count++;
		printf("%d\n", count);
	}
	return 0;
}