HDU2069 硬币母函数DP

最新推荐文章于 2019-08-19 22:59:08 发布

chalstlve

最新推荐文章于 2019-08-19 22:59:08 发布

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分类专栏：编程题文章标签： C

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本文介绍了一种使用母函数的方法来解决硬币找零问题的程序实现。该程序能够找出对于给定金额的所有可能的不同组合数量，使用了包括1美分、5美分、10美分、25美分和50美分在内的五种硬币。

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Suppose there are 5 types of coins: 50-cent, 25-cent, 10-cent, 5-cent, and 1-cent. We want to make changes with these coins for a given amount of money.
For example, if we have 11 cents, then we can make changes with one 10-cent coin and one 1-cent coin, or two 5-cent coins and one 1-cent coin, or one 5-cent coin and six 1-cent coins, or eleven 1-cent coins. So there are four ways of making changes for 11 cents with the above coins. Note that we count that there is one way of making change for zero cent.
Write a program to find the total number of different ways of making changes for any amount of money in cents. Your program should be able to handle up to 100 coins.

母函数经典题

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

const int N=300;

int c1[N][110],c2[N][110];
int res[N],money[6]={0,1,5,10,25,50};

void Init(){
    memset(c1,0,sizeof(c1));
    memset(c2,0,sizeof(c2));
    c1[0][0]=1;
    for(int i=1;i<=5;i++){
        for(int j=0;j<=250;j++)
            for(int k=0;j+k*money[i]<=250;k++)
                for(int p=0;k+p<=100;p++)       //限制硬币总数不超过100
                    c2[j+k*money[i]][p+k]+=c1[j][p];
        for(int j=0;j<=250;j++)
            for(int p=0;p<=100;p++){    //限制硬币总数不超过100
                c1[j][p]=c2[j][p];
                c2[j][p]=0;
            }
    }
    for(int i=1;i<=250;i++)
        for(int j=0;j<=100;j++)
            res[i]+=c1[i][j];
    res[0]=1;       
}

int main(){
    int n;
    Init();
    while(~scanf("%d",&n)){
        printf("%d\n",res[n]);
    }
    return 0;
}