HDU 2069 Coin Change (母函数)

问题描述：找硬币问题，有个限制条件，就是总数不能超过100。思路：将母函数变形，加一维表示数量。代码：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define MID(x,y) ((x+y)>>1)
using namespace std;
typedef long long LL;
int a[5] = {1, 5, 10, 25, 50};
/* generating_function */
const int N = 300;
struct generating_function{
    int c1[N][101], c2[N][101];
    int maxn;
    void init(int n){
        memset(c1, 0, sizeof(c1));
        memset(c2, 0, sizeof(c2));
        c1[0][0] = 1;
        maxn = n;
    }
    void cal(int p);
}ef;
void generating_function::cal(int p){
    int tmp[N][101];
    memset(tmp, 0, sizeof(tmp));
    for (int i = 0; i <= maxn; i ++){
        for (int j = 0; j <= maxn; j += a[p]){
            int k1 = j/a[p];
            if (i + j <= maxn && c2[j][k1])
                for (int k2 = 0; k1+k2 <= 100; k2 ++)
                    tmp[i+j][k1+k2] += c1[i][k2] * c2[j][k1];
        }
    }
    memset(c1, 0, sizeof(c1));
    for (int i = 0; i <= maxn; i ++){
        for (int k = 0; k <= 100; k ++){
            c1[i][k] = tmp[i][k];
        }
    }
    return ;
}
int main(){
    int n;
    while(scanf("%d", &n) == 1){
        ef.init(n);
        for (int i = 0; i < 5; i ++){
            memset(ef.c2, 0, sizeof(ef.c2));
            for (int j = 0; j <= n; j += a[i]){
                ef.c2[j][j/a[i]] = 1;
            }
            ef.cal(i);
        }

        int res = 0;
        for (int k = 0; k <= 100; k ++){
            res += ef.c1[n][k];
        }
        printf("%d\n", res);
    }
	return 0;
}

举杯独醉，饮罢飞雪，茫然又一年岁。 ------AbandonZHANG