Sum of Consecutive Prime Numbers
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 19350 Accepted: 10619
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
题目大意：
一个数可以由若干种连续的素数序列求和得到，比如说41 = 2+3+5+7+11+13 = 11+13+17 = 41
共有三种不同的素数序列求和得到。给你一个数N，求满足N = 连续的素数序列和的方案数
思路：
很简单的题目，但是用普通方法判断素数可能会超时，这里用了筛法求素数的方法直接用数组Prime
判断是否为素数，另开一个数组PrimeNum用来存所有的素数。
最后就是枚举，求得满足的方案数
#include<stdio.h>
#include<string.h>
int Prime[10010],PrimeNum[10010];
int IsPrime()//筛法求素数
{
Prime[0] = Prime[1] = 0;
for(int i = 2; i <= 10000; i++)
Prime[i] = 1;
for(int i = 2; i <= 10000; i++)
{
for(int j = i+i; j <= 10000; j+=i)
Prime[j] = 0;
}
int num = 0;
for(int i = 0; i <= 10000; i++)
if(Prime[i])
PrimeNum[num++] = i;
return num;
}
int main()
{
int num = IsPrime();
int N;
while(~scanf("%d",&N) && N!=0)
{
int count = 0;
for(int i = 0; PrimeNum[i]<=N && i < num; i++)//枚举
{
int sum = 0;
for(int j = i; PrimeNum[j]<=N && j < num; j++)
{
sum += PrimeNum[j];
if(sum == N)
{
count++;
break;
}
if(sum > N)
break;
}
}
printf("%d\n",count);
}
return 0;
}
