hdu 2069 Coin Change

Coin Change

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13471    Accepted Submission(s): 4493

Problem Description

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.

 

Input

The input file contains any number of lines, each one consisting of a number ( ≤250 ) 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

1126
/*题解：    随着做母函数题数的增多，发现用一位数组解母函数问题越来越无力，以后有个数限制的尽量用二维数组吧。     其实有数量限制的题是可以用暴力枚举解的，但仅仅用枚举的话，不仅自己失去了学习母函数的机会，题目也失去了意义，变成简单的水题。     就和解大数问题一样，java都能水过，为什么还用C？因为学的是过程、思想，而不是简简单单要个A！     言归正传，此题用二维数组做(先打好表)，例如：c1[a][b],a指的是钱数，b指的是钱币的个数。     */
#include<cstdio>#include<cstring>int c1[1010][1010],c2[1010][1010];int main(){    int i,j,k,n,h,sum,coin[5]={1,5,10,25,50};        memset(c1,0,sizeof(c1));        memset(c2,0,sizeof(c2));        for(i=0; i<=250; i+=coin[0])    //对第一个括号内的数据初始化         c1[i][i]=1;          for(i=1; coin[i]<=250&&i<=4; i++)        {            for(j=0; j<=250; j++)            {                for(k=0; k+j<=250; k+=coin[i])                {                    for(h=0; h+k/coin[i]<=100; h++)                     c2[k+j][h+k/coin[i]]+=c1[j][h];                }            }            for(j=0; j<=250; j++)            for(h=0; h<=100; h++)            {                c1[j][h]=c2[j][h];                c2[j][h]=0;            }        }    while(scanf("%d",&n)!=EOF)    {        for(i=0,sum=0; i<=100; i++)        sum+=c1[n][i];                 printf("%d\n",sum);    }    return 0;}