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| 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 2001028992079324314
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; }
poj3132 Sum of Different Primes
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原文地址：http://blog.csdn.net/mengzhengnan/article/details/45849393
