uva 357 - Let Me Count The Ways(动态规划-注意dp初始化的问题)

1、http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=293

2、题目大意：给定五种硬币，及一个面值额，看有多少种方法可以组成当前面值额，很简单的dp题目，不过超时了一遍，注意处理这种问题时，dp的初始化只需一遍即可，不需要每次都初始化，后边如果n很大的话，可以用前边用过的dp[][]值，不需要再计算，否则会超时

3、题目

Let Me Count The Ways

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 m ways to produce n cents change.

There is only 1 way to produce n cents change.

Sample input

17
11
4

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.

4、AC代码

#include<stdio.h>
#define N 30005
#define ll long long
#include<string.h>
int v[5]={1,5,10,25,50};
ll dp[7]
;//dp[i][j]表示前i种硬币组成j分的种类数
ll DP(int i,int j)
{
if(dp[i][j]!=-1)
return dp[i][j];
dp[i][j]=0;
for(int k=0;j-k*v[i]>=0;k++)
{
dp[i][j]+=DP(i-1,j-v[i]*k);
}
return dp[i][j];
}
int main()
{
int n;
memset(dp,-1,sizeof(dp));//在外边只初始化一次即可，否则会超时
while(scanf("%d",&n)!=EOF)
{

for(int i=0;i<=n;i++)
dp[0][i]=1;
ll ans=DP(4,n);
if(ans==1)
printf("There is only 1 way to produce %d cents change.\n",n);
else
printf("There are %lld ways to produce %d cents change.\n",ans,n);
}
return 0;
}