POJ3132 Sum of Different Primes
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 |
用素数筛打一个素数表出来,然后在素数表上背包动规。
1 /*by SilverN*/ 2 #include<algorithm> 3 #include<iostream> 4 #include<cstring> 5 #include<cstdio> 6 #include<cmath> 7 using namespace std; 8 int read(){ 9 int x=0,f=1;char ch=getchar(); 10 while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();} 11 while(ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();} 12 return x*f; 13 } 14 const int mxn=5000; 15 int pri[mxn],cnt=0; 16 int vis[mxn]; 17 int n,k; 18 int f[mxn][30]; 19 void Pri(){ 20 int i,j; 21 for(i=2;i<=n;i++){ 22 if(!vis[i]){ 23 pri[++cnt]=i; 24 } 25 for(j=1;j<=cnt && i*pri[j]<=n;j++){ 26 vis[i*pri[j]]=1; 27 if(i%pri[j]==0)break; 28 } 29 } 30 return; 31 } 32 int main(){ 33 n=1250; 34 Pri(); 35 int i,j; 36 f[0][0]=1; 37 for(i=1;i<=cnt;i++){ 38 for(j=n;j>=pri[i];j--){ 39 for(int k=1;k<=15;k++) 40 f[j][k]+=f[j-pri[i]][k-1]; 41 } 42 } 43 while(scanf("%d%d",&n,&k) && n){ 44 printf("%d\n",f[n][k]); 45 } 46 return 0; 47 }