POJ3132――Sum of Different Primes - c++编程基础

Time Limit: 5000MS	?	Memory Limit: 65536K
Total Submissions: 3243	?	Accepted: 2016

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

Sample Output

Source

类似于背包问题，不过有2个代价

设dp[i][j]表示用j个素数表示出i这个数的方案数

dp[i][j] += dp[ i - prime][j - 1]

大家知道01背包优化到一维时体积需要逆序循环是因为 f[i][j] 需要由 f[i - 1][j - c]推得，而f[i - 1][j - c]又要存在dp[j - c]中，如果顺序循环，dp[j - c]中将存放f[i][j - c]，所以就不对了，对于此题，我们其实已经将状态降到二维了（本来是三维），所以同样的，在循环体积时，需要逆序

#include

#include #include #include #include #include #include #include #include #include #include #include using namespace std; __int64 dp[1200][16]; int tot; int prime[1200]; bool valid[1200]; void get_prime() { memset (valid, 0, sizeof(valid)); for (int i = 2; i <= 1120; ++i) { valid[i] = 1; } for (int i = 2; i <= 1120; ++i) { if (valid[i]) { if (1120 / i < i) { break; } for (int j = i * i; j <= 1120; j += i) { valid[j] = 0; } } } for (int i = 2; i <= 1120; ++i) { if (valid[i]) { prime[++tot] = i; } } } int main() { tot = 0; get_prime(); int n, k; memset (dp, 0, sizeof(dp)); dp[0][0] = 1; for (int l = 1; l <= tot; ++l) { for (int j = 1120; j >= 2; --j) { for (int i = 14; i >= 1; --i) { if (j < prime[l]) { continue; } dp[j][i] += dp[j - prime[l]][i - 1]; } } } while (~scanf("%d%d", &n, &k)) { if (!n && !k) { break; } printf("%I64d\n", dp[n][k]); } return 0; }