【01背包方案数】POJ-3132 Sum of Different Primes

Sum of Different Primes

Time Limit: 5000MS Memory Limit: 65536K   

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

————————————————————平凡の分割線————————————————————

思路：将一个数分成k个不同的素数的和。

首先打素数表，然后素数就是01背包问题。

被拆分的数字大小就是背包容量。

除了容量限制，背包还有一个物品个数的限制。

给出两种解法：DP、母函数

以下四种DP方式都对：

【一】

for(int k = 0; k < cnt && p[k] <= v; k++) {for(int j = v; j >= p[k]; j--) {for(int i = 1; i <= n; i++) {dp[j][i] += dp[j-p[k]][i-1];}}}

【二】

for(int k = 0; k < cnt && p[k] <= v; k++) {for(int i = n; i >= 1; i--) {for(int j = p[k]; j <= n; j++) {dp[j][i] += dp[j-p[k]][i-1];}}}

【三】

for(int k = 0; k < cnt && p[k] <= v; k++) {for(int j = v; j >= p[k]; j--) {for(int i = n; i >= 1; i--) {dp[j][i] += dp[j-p[k]][i-1];}}}

【四】

for(int k = 0; k < cnt && p[k] <= v; k++) {for(int i = n; i >= 1; i--) {for(int j = v; j >= p[k]; j--) {dp[j][i] += dp[j-p[k]][i-1];}}}

原因分析：根据状态转移方程，dp[j][i] += dp[j-p[k]][i-1]，只要我们在更新dp[j][i]之前没有更新dp[j-p[k]][i-1]，就没有后效性。通过i、j作出坐标系，根据更新的顺序划线，就可以很容易地发现这件事。

下面给出方式【一】的图：

代码如下：

/*ID: j.sure.1PROG:LANG: C++*//****************************************/#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <ctime>#include <cmath>#include <stack>#include <queue>#include <vector>#include <map>#include <string>#include <climits>#include <iostream>#define LL long longusing namespace std;const int INF = 0x3f3f3f3f;/****************************************/const int N = 1200;bool vis[N];int p[N], dp[1200][20];int get_prime() {    int cnt = 0;    vis[1] = 1;    for(int i = 2; i < N; i++) {        if(!vis[i]) p[cnt++] = i;        for(int j = 0; j < cnt && p[j]*i < N; j++) {            vis[p[j]*i] = 1;            if(i % p[j] == 0) break;        }    }return cnt;}int main(){#ifdef J_Surefreopen("000.in", "r", stdin);freopen("999.out", "w", stdout);#endifmemset(vis, 0, sizeof(vis));int cnt = get_prime();int v, n;while(scanf("%d%d", &v, &n), v||n) {memset(dp, 0, sizeof(dp));dp[0][0] = 1;for(int k = 0; k < cnt && p[k] <= v; k++) {for(int j = v; j >= p[k]; j--) {for(int i = n; i >= 1; i--) {dp[j][i] += dp[j-p[k]][i-1];}}}printf("%d\n", dp[v][n]);}return 0;}

下面给出母函数的解法：

P.S. 由于是01背包，只需要在 j 的循环中，上下限设置为0到1即可。

代码如下：

/*ID: j.sure.1PROG:LANG: C++*//****************************************/#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <ctime>#include <cmath>#include <stack>#include <queue>#include <vector>#include <map>#include <string>#include <climits>#include <iostream>#define LL long longusing namespace std;const int INF = 0x3f3f3f3f;/****************************************/const int N = 1120, M = 14;bool vis[N+5];int p[N+5], c1[N+5][M+5], c2[N+5][M+5];int get_prime() {    int cnt = 0;    vis[1] = 1;    for(int i = 2; i < N+5; i++) {        if(!vis[i]) p[cnt++] = i;        for(int j = 0; j < cnt && p[j]*i < N+5; j++) {            vis[p[j]*i] = 1;            if(i % p[j] == 0) break;        }    }return cnt;}int main(){#ifdef J_Surefreopen("000.in", "r", stdin);freopen("999.out", "w", stdout);#endifmemset(vis, 0, sizeof(vis));int cnt = get_prime();memset(c1, 0, sizeof(c1));c1[0][0] = 1;for(int i = 0; i < cnt; i++) {memset(c2, 0, sizeof(c2));for(int j = 0; j <= 1 && j*p[i] <= N; j++) {for(int k = 0; k + j*p[i] <= N; k++) {for(int l = 0; l + j <= M; l++) {c2[k+j*p[i]][l+j] += c1[k][l];}}}memcpy(c1, c2, sizeof(c1));}int x, y;while(scanf("%d%d", &x, &y), x||y) {printf("%d\n", c1[x][y]);}return 0;}