Sum of Different Primes
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 3293 | Accepted: 2052 |
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 kdifferent 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
先打表找出所有的质数，再转化为与背包问题类似的问题，装态转移方法很简单，dp[i][j] += dp[i-pri][j-1];
dp[i][j]表示i分解成j的个数，pri表示枚举质数，即可解！
// ConsoleApplication1.cpp : 定义控制台应用程序的入口点。
//
#include "stdafx.h"
#include "stdio.h"
#include "string.h"
#include "math.h"
#include <algorithm>
#include <vector>
#include <queue>
#include <iostream>
using namespace std;
#define N 1125
#define SCANF scanf_s
bool pri[N];
int prime[200];
int dp[N][15];
int prinum = 0;
int getPrimeNumMid(int sum)
{
int left = 1, right = prinum - 1, mid;
while (left < right - 1)
{
mid = (left + right) / 2;
if (prime[mid] > sum)
{
right = mid;
}
else if (prime[mid] < sum)
{
left = mid;
}
else
{
return mid;
}
}
if (prime[right] <= sum)
return right;
else
return left;
}
void init(int n,int knum)
{
memset(pri, true, sizeof(pri));
pri[0] = pri[1] = false;
for (int i = 2; i < sqrt((float)N); i++)
{
for (int j = i + i; j < N; j += i)
{
pri[j] = false;
}
}
prinum = 1;
for (int i = 0; i < N; i++)
{
if (pri[i])
{
prime[prinum] = i;
prinum++;
}
}
int pnum = getPrimeNumMid(n);
//printf("%d\n", prime[pnum]);
for (int i = 0; i <= n; i++)
{
for (int j = 0; j <= knum; j++)
{
dp[i][j] = 0;
}
}
dp[0][0] = 1;
for (int k = 1; k <= pnum; k++)
{
for (int i = n; i >= 0; i--)
{
for (int j = 1; j <= knum; j++)
{
if (i - prime[k] >= 0)
dp[i][j] += dp[i - prime[k]][j - 1];
//printf("%d %d %d \n", i, j, dp[i][j]);
}
}
}
}
int main()
{
int n, knum;
init(1120,14);
while (SCANF("%d%d", &n , &knum) != EOF)
{
if (n == 0 && knum == 0)
break;
printf("%d\n", dp[n][knum]);
}
return 0;
}