uva674 - Coin Change(dp)

题目大意是用 1,5,10,25,50 五种钱组成一个给定的N ，求有多少种组发。。

dp[i]为 价值I 的个数， 那么dp[i] += dp[i-k]  (k为上面五种钱币 。。 ）

 

题目：

Coin Change 

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, two 5-cent coins and one 1-cent coin, 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 7489 cents.

 

Input 

The input file contains any number of lines, each one consisting of a number for the amount of money in cents.

 

Output 

For each input line, output a line containing the number of different ways of making changes with the above 5 types of coins.

 

Sample Input 

 

11
26

 

Sample Output 

4
13

 

 

代码：

#include <iostream>
#include <memory.h>
using namespace std;

int dp[10000];
int M[5] = {50,25,10,5,1};
int main()
{
    int total;
    while(cin>>total)
    {
        if(total ==0){ cout<<"0"<<endl;continue;}

        dp[0]=1;
            for(int j = 0;j<=total;j++)
            {
                for(int i=0;i<5;i++)
                {
                    dp[j+M[i]] += dp[j];
                }
        }
        cout<<dp[total]<<endl;
        memset(dp,0,sizeof(dp));
    }


    return 0;
}

 

 

代码2

：

 

#include <iostream>
#include <memory.h>
using namespace std;

int dp[10000][6];
int M[5] = {1,5,10,25,50};
int main()
{
    int total;


        dp[0][0]=1;

        for(int i=1;i<=8000;i++)
        {
            for(int j=0;j<5;j++)
            {
                for(int k=0;k<=j;k++)　　　　//一开始这里写错了 k 必须小于j 否则会出现多加的情况
                {
                    if( i<M[j]  )break;
                    dp[i][j] += dp[i-M[j]][k];
                }
            }
        }

    while(cin>>total)
    {
        if(total ==0){ cout<<"0"<<endl;continue;}

        int ans=0;
        for(int i=0;i<5;i++)
        {
            ans+=dp[total][i];
        }
        cout<<ans<<endl;

    }


    return 0;
}