UVA 357 Let Me Count The Ways 动态规划解法、母函数解法

Description

After making a purchase at a large department store, Mel's change was 17 cents. He received 1 dime, 1 nickel, and 2 pennies. Later that day, he was shopping at a convenience store. Again his change was 17 cents. This time he received 2 nickels and 7 pennies. He began to wonder ' "How many stores can I shop in and receive 17 cents change in a different configuration of coins? After a suitable mental struggle, he decided the answer was 6. He then challenged you to consider the general problem.

Write a program which will determine the number of different combinations of US coins (penny: 1c, nickel: 5c, dime: 10c, quarter: 25c, half-dollar: 50c) which may be used to produce a given amount of money.

Input

The input will consist of a set of numbers between 0 and 30000 inclusive, one per line in the input file.

Output

The output will consist of the appropriate statement from the selection below on a single line in the output file for each input value. The number m is the number your program computes, n is the input value.

There are mways to produce ncents change.

There is only 1 way to produce ncents change.

Sample input

17 114

Sample output

There are 6 ways to produce 17 cents change. There are 4 ways to produce 11 cents change. There is only 1 way to produce 4 cents change.

可以作为一个有关钱币兑换的模板题。

DP

#include<stdio.h>#include<string.h>int v[5]={1,5,10,25,50};long long dp[30009]; //定义为long long形式，避免溢出 int main(){int i,j;long long n;while(~scanf("%lld",&n)){memset(dp,0,sizeof(dp));dp[0]=1; for(i=0;i<5;i++)    for(j=v[i];j<=n;j++)        dp[j]+=dp[j-v[i]];  //状态转移方程 if(dp[n]==1)   //区分等一和不等一时的输出 printf("There is only 1 way to produce %lld cents change.\n",n);elseprintf("There are %lld ways to produce %lld cents change.\n",dp[n],n);}return 0;}

母函数

#include<iostream>using namespace std;long long  c1[30001];long long  c2[30001];int v[5]={1,5,10,25,50};int fun()  //先全部计算出来，避免超时 {long long i,j,k;for(i=0;i<=30001;i++){c1[i]=1;c2[i]=0;}for(i=1;i<5;i++){for(j=0;j<=30001;j++)  for(k=0;k+j<=30001;k+=v[i])   c2[k+j]+=c1[j];   for(j=0;j<=30001;j++)   {   c1[j]=c2[j];   c2[j]=0;   }}}int main(){    long long n;fun();while(~scanf("%lld",&n)){    if(c1[n]==1)   printf("There is only 1 way to produce %lld cents change.\n",n);elseprintf("There are %lld ways to produce %lld cents change.\n",c1[n],n);}return 0;}