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1、
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
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.
The input will consist of a set of numbers between 0 and 30000 inclusive, one per line in the input file.
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.
17 114
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#define N 30005#define ll long long#include int v[5]={1,5,10,25,50};ll dp[7][N];//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;}
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