UVa 674 Coin Change（经典DP）

题意：

有1,5,15,25,50五种货币，问n元最多有多少种兑换方法

思路：

dp[i]表示i钱最多的兑换方法，dp[i+num[k]] += dp[i]。此时要注意dp的转移顺序。需要仔细理解。

（一开始用母函数做，果断超时了。也可以dfs记忆化搜索）

View Code

#include <cstdio>

#include <cstdlib>

#include <cstring>



const int MAXN = 8000;



long long int dp[MAXN];

int a[5] = {1, 5, 10, 25, 50};



int main()

{

    int n;

    while (scanf("%d", &n) != EOF)

    {

        memset(dp, 0, sizeof(dp));

        dp[0] = 1;



        for (int i = 0; i < 5; ++i)

            for (int j = 0; j <= n; ++j)

                dp[j+a[i]] += dp[j];



        printf("%lld\n", dp[n]);

    }

    return 0;

}

母函数（超时）：

View Code

#include <cstdio>

#include <cstdlib>

#include <cstring>



const int MAXN = 8000;



long long int c1[MAXN], c2[MAXN];



int main()

{

    int n;

    while (scanf("%d", &n) != EOF)

    {

        for (int i = 0; i <= n; ++i)

            c1[i] = 1, c2[i] = 0;



        int num[5] = {5, 10, 25, 50};



        for (int i = 0; i < 4; ++i)

        {

            for (int j = 0; j <= n; ++j)

                for (int k = 0; k + j <= n; k += num[i])

                    c2[k+j] += c1[j];

            for (int j = 0; j <= n; ++j)

                c1[j] = c2[j], c2[j] = 0;

        }

        printf("%lld\n", c1[n]);

    }

    return 0;

}