POJ3132Sum of Different Primes題解動态規劃DP

Sum of Different Primes

Time Limit: 5000MS	Memory Limit: 65536K
Total Submissions: 2160	Accepted: 1334

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, 19} 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 簡單的二維費用01背包 狀态： d[i][j]表示j個素數組成和為i的方法數 狀态轉移方程： d[k][j]+=d[k-p[i]][j-1];

邊界： d[0][0]=1; 代碼： #include<cstdio>

#define N 1200

int p[N];

bool f[N];

void init()

{

int i,j,m=0;

f[0]=f[1]=1;

for(i=4;i<=N;i+=2)

f[i]=1;

for(i=3;i*i<=N;i+=2)

if(!f[i])

for(j=i*i;j<=N;j+=i*2)

f[j]=1;

for(i=2;i<=N;i++)

if(!f[i])

p[m++]=i;

}

int main()

{

init();

int n,m;

while(scanf("%d%d",&n,&m),n+m)

{

int i,j,k,d[N][15]={0};

d[0][0]=1;

for(i=0;p[i]<=n;i++)

for(k=n;k>=p[i];k--)

for(j=m;j>0;j--)

d[k][j]+=d[k-p[i]][j-1];

printf("%d/n",d[n][m]);

}

}