最左/右边的数-hd1060/1061

左边：

http://acm.hdu.edu.cn/showproblem.php?pid=1060

 1 #include <iostream>
 2 #include <math.h>
 3 using namespace std;
 4 
 5 int main()
 6 {
 7     int n,m,d,i;
 8     double l;
 9     while(cin>>n)
10 
11     {
12         for(i=1;i<=n;i++)
13         {
14             cin>>m;
15             l=m*log10(m)-(long long)(m*log10(m));
16             d=(int)pow(10,l);
17             cout<<d<<endl;
18         }
19     }
20     return 0;
21 }

View Code

 1 #include <stdio.h>
 2 #include <stdlib.h>
 3 #include <math.h>
 4 int main()
 5 {
 6     int m,i,n;
 7     double l,d;
 8     while(~scanf("%d",&m))
 9     {
10         for(i=1;i<=m;i++)
11         {
12             scanf("%d",&n);
13             l=n*log10(n)-(long long)(n*log10(n));
14             d=pow(10,l);
15             printf("%d
",(int)d);
16         }
17     }
18     return 0;
19 }

View Code

右边：

http://acm.hdu.edu.cn/showproblem.php?pid=1061

 1 #include <iostream>
 2 
 3 using namespace std;
 4 
 5 int main()
 6 {
 7     int n,m,i,j,k,data[10][4]={{0,0,0,0},{1,1,1,1},{6,2,4,8},{1,3,9,7},{6,4,6,4},{5,5,5,5},{6,6,6,6},{1,7,9,3},{6,8,4,2},{1,9,1,9}};
 8     while(cin>>n)
 9     {
10         for(i=1;i<=n;i++)
11         {
12             cin>>m;
13             k=m%10;
14             j=m%4;
15             cout<<data[k][j]<<endl;
16         }
17     }
18     return 0;
19 }

View Code

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29 View Code
30 
31 Problem : 1061 ( Rightmost Digit )     Judge Status : Accepted
32 RunId : 16073921    Language : C    Author : wangyawei
33 Code Render Status : Rendered By HDOJ C Code Render Version 0.01 Beta
34 #include <stdio.h>
35 #include <stdlib.h>
36 #include<string.h>
37 int main()
38 {
39      int data[10][4]={{0},{1},{6,2,4,8},{1,3,9,7},{6,4},{5},{6},{1,7,9,3},{6,8,4,2},{1,9}};
40      int n,i,t,num;
41      while(~scanf("%d",&num))
42      {
43          for(i=1;i<=num;i++)
44          {
45              scanf("%d",&n);
46          t=n%10;
47          if(t==0||t==1||t==5||t==6)
48          printf("%d
",t);
49          else if(t==4||t==9)
50          printf("%d
",data[t][n%2]);
51          else if(t==2||t==3||t==7||t==8)
52          printf("%d
",data[t][n%4]);
53          }
54      }
55     return 0;
56 }

View Code