uva 147 Dollars

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=83

147 - Dollars

Time limit: 3.000 seconds

Dollars

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

 

 

分析：

這個題跟UVA674題相似，只是這個題有更多的錢的種類，而且須要用long long來保存

作的方法是先乘以100消除浮點數的偏差，而後計算，須要注意的是輸出格式有要求，有特殊的空格要求~

 

AC代碼：

 

遞推：

 

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 using namespace std;
 5 const int maxn=30001;
 6 int num[12]={0,5,10,20,50,100,200,500,1000,2000,5000,10000};
 7 int a,b;
 8 long long f[maxn][12];
 9 void Init()
10 {
11     for(int i=0;i<12;i++)
12     f[0][i]=1;
13     for(int i=1;i<maxn;i++)
14     for(int j=1;j<12;j++)
15         for(int k=0;num[j]*k<=i;k++)
16         f[i][j]+=f[i-num[j]*k][j-1];
17 }
18 int main()
19 {
20     Init();
21     while(scanf("%d.%d",&a,&b)&&(a+b))
22     {
23     int n=a*100+b;
24     printf("%6.2lf%17lld\n",n*1.0/100,f[n][11]);
25     }
26     return 0;
27 }

View Code

 

 

揹包：

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 using namespace std;
 5 const int maxn=30001;
 6 int num[12]={0,5,10,20,50,100,200,500,1000,2000,5000,10000};
 7 int a,b;
 8 long long f[maxn][12];
 9 void Init()
10 {
11     for(int i=0;i<12;i++)
12     f[0][i]=1;
13     for(int i=1;i<maxn;i++)
14     for(int j=1;j<12;j++)
15         for(int k=0;num[j]*k<=i;k++)
16         f[i][j]+=f[i-num[j]*k][j-1];
17 }
18 int main()
19 {
20     Init();
21     while(scanf("%d.%d",&a,&b)&&(a+b))
22     {
23     int n=a*100+b;
24     printf("%6.2lf%17lld\n",n*1.0/100,f[n][11]);
25     }
26     return 0;
27 }

View Code

 

後來想到能夠再除以5（題目說是5的倍數），來減小運算量，提升運算效率。

 

 

 

 1 #include <cstdio>
 2 using namespace std;
 3 
 4 long long f[30010];
 5 const int num[] = {2000, 1000, 400, 200, 100, 40, 20, 10, 4, 2, 1};
 6 
 7 int main()
 8 {
 9     f[0] = 1;
10     for(int i = 0 ;i <= 10;i ++)
11         for(int j = num[i] ;j <= 6005;j ++)
12             f[j] += f[j - num[i]];
13     double x;
14     while(scanf("%lf", &x) , x)
15     {
16         double tx = x * 20;
17         printf("%6.2lf%17lld\n", x , f[(int)tx] );
18     }
19     return 0;
20 }