hdu 1028 Ignatius and the Princess III 完全背包

      这题以前是母函数做的，今天看了DD的背包九讲，该用背包模型做。

      这题是完全背包。设dp[i][v] 为用前i个数组成v的方案数，状态方程如下：dp[i][v] = sum(dp[i-1][v-k*i])

 

#include <iostream>
using namespace std;

const int N = 122;
//dp[i][v] 为用前i个数组成v的方案数
//dp[i][v] = sum(dp[i-1][v-k*i])
int dp[N][N];

int main()
{
	int n;

	while (cin >> n)
	{
		memset(dp, 0, sizeof(dp));
		dp[0][0] = 1;
		
		for (int i = 1; i <= n; i++)		
			for (int k = 0; k*i <= n; k++)
				for (int j = k*i; j <= n; j++)	
					dp[i][j] += dp[i-1][j-k*i];
									
		cout << dp[n][n] << endl;
	}
	return 0;
}

 

可以优化成一维数组

for (int i = 1; i <= n; i++)
  for (int j = i; j <= n; j++)
     dp[j] += dp[j-i];