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PKU3121 Sum of Different Primes

Posted on 2007-02-18 10:02 oyjpart 阅读(1183) 评论(2)  编辑 收藏 引用

Sum of Different Primes
Time Limit:5000MS  Memory Limit:65536K
Total Submit:362 Accepted:219

Description

A positive integer may be expressed as a sum of different prime numbers (primes), in one way or another. Given two positive integers n and k, you should count the number of ways to express n as a sum of k different primes. Here, two ways are considered to be the same if they sum up the same set of the primes. For example, 8 can be expressed as 3 + 5 and 5 + 3 but the are not distinguished.

When n and k are 24 and 3 respectively, the answer is two because there are two sets {2, 3, 18} and {2, 5, 17} whose sums are equal to 24. There are not other sets of three primes that sum up to 24. For n = 24 and k = 2, the answer is three, because there are three sets {5, 19}, {7, 17} and {11, 13}. For n = 2 and k = 1, the answer is one, because there is only one set {2} whose sum is 2. For n = 1 and k = 1, the answer is zero. As 1 is not a prime, you shouldn’t count {1}. For n = 4 and k = 2, the answer is zero, because there are no sets of two different primes whose sums are 4.

Your job is to write a program that reports the number of such ways for the given n and k.

Input

The input is a sequence of datasets followed by a line containing two zeros separated by a space. A dataset is a line containing two positive integers n and k separated by a space. You may assume that n ≤ 1120 and k ≤ 14.

Output

The output should be composed of lines, each corresponding to an input dataset. An output line should contain one non-negative integer indicating the number of the ways for n and k specified in the corresponding dataset. You may assume that it is less than 231.

Sample Input

24 3 
24 2 
2 1 
1 1 
4 2 
18 3 
17 1 
17 3 
17 4 
100 5 
1000 10 
1120 14 
0 0

Sample Output

2 
3 
1 
0 
0 
2 
1 
0 
1 
55 
200102899 
2079324314

Source
Japan 2006

如何写无重复的情况呢?
刚开始的时候我写的是按以前写搜索的那种写法 加了最大数的限制
但是数组多了一维 后来想起来其实可以这样写 现在居然忘记了。。faint

Solution
//by oyjpArt
int n, s; //全数,阶段
int st[MAXN][MAXS];
bool test[MAXN]; //这个是删数法的规则
int p[200];
int np;

void pre()
{
 int i, j, k;
 memset(test, true, sizeof(test));
 memset(st, 0, sizeof(st));
 int np = 0;
 for(i=2; i<MAXN; i++)
  if(test[i])
  {
   p[np++] = i;
   for(j=i+i; j<MAXN; j+=i)
    test[j] = 0;
  }
 st[0][0] = 1;
 for(i=0; i<np; i++) //阶段
  for(j=1120-p[i]; j>=0; j--)
   for(k = 14; k>=1; k--)
    st[j+p[i]][k] += st[j][k-1];
}
int main()
{
 pre();
 while(scanf("%d%d", &n, &s), n>0)
 {
  printf("%d\n", st[n][s]);
 }
 return 0;
}

Feedback

# re: PKU3121 Sum of Different Primes   回复  更多评论   

2008-07-01 18:13 by ssadwll
自己都没交成功

# re: PKU3121 Sum of Different Primes   回复  更多评论   

2008-07-01 18:43 by oyjpart
恩?

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