UVA - 1213
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 andk, When n and k are 24 and 3 respectively, the answer is two because there are two sets {2, Your job is to write a program that reports the number of such ways for the given n and k. InputThe 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 OutputThe 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 ways for n and k specified Sample Input24 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 Output2 3 1 0 0 2 1 0 1 55 200102899 2079324314 Source Root :: AOAPC II: Beginning Algorithm Contests (Second Edition) (Rujia Liu) :: Chapter 10. Maths :: Exercises Root :: Competitive Programming 2: This increases the lower bound of Programming Contests. Again (Steven & Felix Halim) :: Problem Solving Paradigms :: Dynamic Programming :: 0-1 Root :: Competitive Programming 3: The New Lower Bound of Programming Contests (Steven & Felix Halim) :: Problem Solving Paradigms :: Dynamic Programming :: 0-1 |
1120 and k题意是选择k个质数使其和为n，先搞一个素数表然后dp，dp[i][j]表示选了j个数和位i的方案数。
#include <bits/stdc++.h>
#define foreach(it,v) for(__typeof((v).begin()) it = (v).begin(); it != (v).end(); ++it)
using namespace std;
typedef long long ll;
const int maxn = 1120;
bool check[maxn];
int prime[1120];
ll dp[1122][15];
int init(int n)
{
memset(check,0,sizeof check);
int tot = 0;
check[0] = check[1] = 1;
for(int i = 2; i <= n; i++) {
if(!check[i])prime[tot++] = i;
for(int j = 0; j < tot; ++j) {
if((ll)i*prime[j]>n)break;
check[i*prime[j]] = true;
if(i%prime[j]==0)break;
}
}
return tot;
}
int main()
{
int n,k;
int tot = init(maxn-1);
while(~scanf("%d%d",&n,&k)&&n) {
memset(dp,0,sizeof dp);
dp[0][0] = 1;
for(int i = 0; prime[i]<=n&&i < tot; i++) {
int limt = min(i+1,k);
int w = prime[i];
for(int V = n; V >= w; V--)
for(int j = 1; j <= limt; j++)if(dp[V-w][j-1]){
dp[V][j] += dp[V-w][j-1];
}
}
cout<<dp[n][k]<<endl;
}
return 0;
}