BNUOJ 17286 Dollars

Dollars

Time Limit: 3000ms

Memory Limit: 131072KB

This problem will be judged on  UVA. Original ID: 147
64-bit integer IO format: %lld      Java class name: Main

 

New Zealand currency consists of $100, $50, $20, $10, and $5 notes and $2, $1, 50c, 20c, 10c and 5c coins. Write a program that will determine, for any given amount, in how many ways that amount may be made up. Changing the order of listing does not increase the count. Thus 20c may be made up in 4 ways: 1  20c, 2  10c, 10c+2  5c, and 4  5c.

 

Input

Input will consist of a series of real numbers no greater than $300.00 each on a separate line. Each amount will be valid, that is will be a multiple of 5c. The file will be terminated by a line containing zero (0.00).

 

Output

Output will consist of a line for each of the amounts in the input, each line consisting of the amount of money (with two decimal places and right justified in a field of width 6), followed by the number of ways in which that amount may be made up, right justified in a field of width 17.

 

Sample input

 

0.20

2.00

0.00

 

Sample output

 

  0.20                4

  2.00              293


解题：几个坑。一是要用long long ,而是要输入0 终止程序

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 #include <cmath>

 5 #include <algorithm>

 6 #include <climits>

 7 #include <vector>

 8 #include <queue>

 9 #include <cstdlib>

10 #include <string>

11 #include <set>

12 #include <stack>

13 #define LL long long

14 #define pii pair<int,int>

15 #define INF 0x3f3f3f3f

16 using namespace std;

17 int c[] = {5,10,20,50,100,200,500,1000,2000,5000,10000};

18 LL dp[30010];

19 int main() {

20     int val = 30000;

21     double ans;

22     memset(dp,0,sizeof(dp));

23     dp[0] = 1;

24     for(int i = 0; i < 11; i++){

25             for(int j = c[i]; j <= val; j++){

26                 dp[j] += dp[j-c[i]];

27             }

28         }

29     while(~scanf("%lf",&ans) && ans != 0.00 ){

30         printf("%6.2f%17lld\n",ans,dp[int(ans*100+0.5)]);

31     }

32     return 0;

33 }

View Code