USACO 2.2 Preface Numbering

TASK: preface
LANG: C++

Compiling...
Compile: OK

Executing...
   Test 1: TEST OK [0.000 secs, 3024 KB]
   Test 2: TEST OK [0.000 secs, 3024 KB]
   Test 3: TEST OK [0.000 secs, 3024 KB]
   Test 4: TEST OK [0.000 secs, 3024 KB]
   Test 5: TEST OK [0.000 secs, 3024 KB]
   Test 6: TEST OK [0.000 secs, 3024 KB]
   Test 7: TEST OK [0.000 secs, 3024 KB]
   Test 8: TEST OK [0.000 secs, 3024 KB]

All tests OK.
 1 /*
 2 ID: jiafeim1
 3 PROG: preface
 4 LANG: C++
 5 */
 6 #include <iostream>
 7 #include <fstream>
 8 #include <algorithm>
 9 using namespace std;
10 
11 #define maxN(x,y) ((x)>(y)?(x):(y))
12 
13 int res[8] = {0};
14 int ge[12][11] = {0};
15 int main()
16 {
17     ofstream fout ("preface.out");
18     ifstream fin ("preface.in");
19 
20     ge[1][0] = 1;
21     ge[2][0] = 2;
22     ge[3][0] = 3;
23     ge[4][0] = 1;
24     ge[4][1] = 1;
25     ge[5][1] = 1;
26     ge[6][0] = 1;
27     ge[6][1] = 1;
28     ge[7][0] = 2;
29     ge[7][1] = 1;
30     ge[8][0] = 3;
31     ge[8][1] = 1;
32     ge[9][0] = 1;
33     ge[9][2] = 1;
34     char str[19] = "IVXLCDM";
35 //              0123456
36     int n ;
37     fin>>n;
38     for(int i = 1;i<=n;++i)
39     {   
40         int g = i %10;
41         int s = (i/10)%10;
42         int b = (i/100)%10;
43         int q = i /1000;
44         res[0]+=ge[g][0];
45         res[1]+=ge[g][1];
46         res[2]+=ge[g][2]+ge[s][0];
47         res[3]+=ge[s][1];
48         res[4]+=ge[s][2]+ge[b][0];
49         res[5]+=ge[b][1];
50         res[6]+=ge[b][2]+ge[q][0];
51     }
52 
53     for(int i = 0;i<=6;++i)
54     {
55         if(res[i]!=0)
56         {
57             fout<<str[i]<<" "<<res[i]<<endl;
58         }
59     }
60     fin.close();
61     fout.close();
62     return 0;
63 }
posted @ 2011-05-04 21:27  幻魇  阅读(206)  评论(0)  编辑  收藏  举报