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];