HDUOJ----Coin Change

Coin Change

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10590    Accepted Submission(s): 3535

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

11

26

 

 

Sample Output

4

13

 

 

Author

Lily

 

 

Source

浙江工业大学网络选拔赛

 

思路： 此题可以采取dfs，但是用分治法还是可以的，优化一下可以达到15ms......

在此贴出代码：

 1 #include<iostream>

 2 using namespace std;

 3 int main()

 4 {

 5     int n,count;

 6     int j,k,m,g,l;

 7     while(cin>>n)

 8     {

 9         count=0;

10         for( j=0;j<=n/50;j++)   //50

11         {  

12             k=m=g=l=0;

13             if(n==j*50+k*25+m*10+g*5+l)

14                         {

15                            count++;

16                            break;

17                         }

18                         else

19                             if(n<j*50+k*25+m*10+g*5+l)

20                                 break;

21 

22             for( k=0;k<=n/25;k++)   //25

23             {

24                 m=g=l=0;

25                 if(n==j*50+k*25+m*10+g*5+l)

26                         {

27                            count++;

28                            break;

29                         }

30                         else

31                             if(n<j*50+k*25+m*10+g*5+l)

32                                 break;

33                 for( m=0;m<=n/10;m++)   //10

34                 {

35                     g=l=0;

36                     if(n==j*50+k*25+m*10+g*5+l)

37                         {

38                            count++;

39                            break;

40                         }

41                         else

42                             if(n<j*50+k*25+m*10+g*5+l)

43                                 break;

44                     for( g=0;g<=n/5;g++)  //5

45                     {

46                         l=0;

47                         if(n==j*50+k*25+m*10+g*5+l)

48                         {

49                            count++;

50                            break;

51                         }

52                         else

53                             if(n<j*50+k*25+m*10+g*5+l)

54                                 break;

55                       for( l=0;l<=100-j-k-m-g;l++)  //1

56                       {

57                         if(n==j*50+k*25+m*10+g*5+l)

58                         {

59                            count++;

60                            break;

61                         }

62                         else

63                             if(n<j*50+k*25+m*10+g*5+l)

64                                 break;

65                        }

66                     }

67                 }

68             }

69         }

70     

71     cout<<count<<endl;

72     }

73 return 0;

74 }

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