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;
}