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.
The input file contains any number of lines, each one consisting of a number for the amount of money in cents.
For each input line, output a line containing the number of different ways of making changes with the above 5 types of coins.
11 26
4 13
題目大意：
給定1 5 10 25 50五種硬幣，給定一個(gè)數(shù)，問(wèn)用這些硬幣有多少種不同的組合方式？
想想好像可以是一個(gè)方程組求解的問(wèn)題嘛：
x1+5*x2+10*x3+25*x4+50*x5=x0;
給定x0，一組滿足條件的x>=0就是解集。啊哈哈哈。可以高斯消元法咯。
不過(guò)，明顯的計(jì)數(shù)問(wèn)題：
dp[i][j]表示i可以由前j種硬幣表示的方法數(shù)。
dp[i][j]=dp[i][j-1]+dp[i-coin[j]][j] i-coin[j]>0
dp[i][j]=dp[i][j-1]+1 i-coin[j]=0;
dp[i][j]=dp[i][j-1] i-coin[j]<0
總結(jié)：
方程類問(wèn)題，一定要先把方程寫(xiě)清楚！！！
#include<stdio.h>
#include<string.h>
#include<math.h>
long long dp[7500][5],coin[5]={1,5,10,25,50};
int GetAns()
{
int i,j;
memset(dp,0,sizeof(dp));
dp[1][0]=dp[1][1]=dp[1][2]=dp[1][3]=dp[1][4]=1;
for (i=2;i<=7489 ;i++ )
{
dp[i][0]=1;
for (j=1;j<5 ;j++ )
if (i-coin[j]>0)
dp[i][j]=dp[i][j-1]+dp[i-coin[j]][j];
else
if (i-coin[j]==0)
dp[i][j]=dp[i][j-1]+1;
else
dp[i][j]=dp[i][j-1];
}
}
int main()
{
int n;
GetAns();
while (scanf("%d",&n)==1)
printf("%lld\n",dp[n][4]);
return 0;
}