HDOJ2069

Coin Change

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 17881    Accepted Submission(s): 6157

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

 

Sample Output

413

 

母函数的表形，多了一维的硬币个数限制。

#include<stdio.h> #include<string.h>int c1[251][101],c2[251][101];//第一维表示钱数，第二维表示使用的硬币的数目int Sum[251];int coin[6]={0,1,5,10,25,50};int main(void){int N;c1[0][0]=1;for(int i=1;i<=5;i++){for(int j=0;j<=250;j++){for(int k=0;k*coin[i]+j<=250;k++){for(int p=0;p+k<=100;p++)//多嵌套了一层记录硬币个数 {c2[k*coin[i]+j][p+k]+=c1[j][p];}}}for(int j=0;j<=250;j++){for ( int k = 0; k != 101; ++ k )             <span style="white-space:pre"></span>{                <span style="white-space:pre"></span>  c1[j][k] = c2[j][k];                <span style="white-space:pre"></span>  c2[j][k] = 0;          <span style="white-space:pre"></span>}}} for ( int j = 0; j != 251; ++ j ) {            for ( int i = 0; i != 101; ++ i )            {               Sum[j] += c1[j][i] ;            }         }     while(scanf("%d",&N)!=EOF){     printf("%d\n",Sum[N]);}return 0;}

也可以用暴力方法

#include<stdio.h>int main(){    int a,b,c,d,e,count,n;    while(scanf("%d",&n)!=EOF)    {        count=0;        for(a=0;a<=100;a++)            for(b=0;5*b<=n;b++)                for(c=0;10*c<=n;c++)                    for(d=0;25*d<=n;d++)                        for(e=0;50*e<=n;e++)                        if(a+5*b+10*c+25*d+50*e==n&&a+b+c+d+e<=100)count++;       printf("%d\n",count);    }}