经典DP：固定钱能换多少次+硬币个数限制(2069)

Coin Change

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16932    Accepted Submission(s): 5802

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

/*------------------Header Files------------------*/#include <iostream>#include <cstring>#include <string>#include <cstdio>#include <algorithm>#include <cstdlib>#include <ctype.h>#include <cmath>#include <stack>#include <queue>#include <deque>#include <map>#include <vector>#include <limits.h>using namespace std;/*------------------Definitions-------------------*/#define LL long long#define PI acos(-1.0)#define INF 0x3F3F3F3F#define MOD 10E9+7#define MAX 500050/*---------------------Work-----------------------*/int dp[500][200];int coin[]={0,50,25,10,5,1};void work(){dp[0][0]=1;for(int i=1;i<=5;i++)for(int j=0;j<=250;j++)for(int k=0;k<100;k++)if(coin[i]+j<=250)dp[coin[i]+j][k+1]+=dp[j][k];else break;int n;while(scanf("%d",&n)==1){int sum=0;for(int i=0;i<=100;i++)sum+=dp[n][i];printf("%d\n",sum);}}/*------------------Main Function------------------*/int main(){//freopen("test.txt","r",stdin);//freopen("cowtour.out","w",stdout);//freopen("cowtour.in","r",stdin);work();return 0;}