So easy

Problem Description
Small W gets two files. There are n integers in each file. Small W wants to know whether these two files are same. So he invites you to write a program to check whether these two files are same. Small W thinks that two files are same when they have the same integer set.
For example file A contains (5,3,7,7),and file B contains (7,5,3,3). They have the same integer set (3,5,7), so they are same.
Another sample file C contains(2,5,2,5), and file D contains (2,5,2,3).
The integer set of C is (2,5),but the integer set of D is (2,3,5),so they are not same.
Now you are expected to write a program to compare two files with size of n.
 
Input
Multi test cases (about 100). Each case contain three lines. The first line contains one integer n represents the size of file. The second line contains n integers a1,a2,a3,,an - represents the content of the first file. The third line contains n integers b1,b2,b3,,bn - represents the content of the second file.
Process to the end of file.
1n100
1ai,bi1000000000
 
Output
For each case, output "YES" (without quote) if these two files are same, otherwise output "NO" (without quote).
 
Sample Input
3
1 1 2
1 2 2
4
5 3 7 7
7 5 3 3
4
2 5 2 3
2 5 2 5
3
1 2 3
1 2 4
 
Sample Output
YES
YES
NO
NO
 
  1 #include <stdio.h> 
  2 
  3 int main(){
  4     int n;
  5     int number;
  6     int i;
  7     int j;
  8     int array1[101];
  9     int array2[101];
 10     int length1;
 11     int length2;
 12     int temp;
 13 
 14     while(scanf("%d",&n)!=EOF){
 15         length1=0;
 16         for(i=0;i<n;i++){
 17             scanf("%d",&number);
 18 
 19             if(i==0){
 20                 array1[0]=number;
 21                 length1++;
 22                 continue;
 23             }
 24 
 25             for(j=0;j<length1;j++){
 26                 if(number==array1[j])
 27                     break;
 28             }
 29 
 30             if(j==length1){
 31                 array1[length1]=number;
 32                 length1++;
 33             }
 34         }
 35 
 36         for(i=0;i<length1-1;i++){
 37             for(j=i+1;j<length1;j++){
 38                 if(array1[i]>array1[j]){
 39                     temp=array1[i];
 40                     array1[i]=array1[j];
 41                     array1[j]=temp;
 42                 }
 43             }
 44         }
 45 
 46         /*for(i=0;i<length1;i++)
 47             printf("%d ",array1[i]);
 48         printf("\n");*/
 49 
 50         length2=0;
 51         for(i=0;i<n;i++){
 52             scanf("%d",&number);
 53 
 54             if(i==0){
 55                 array2[0]=number;
 56                 length2++;
 57                 continue;
 58             }
 59 
 60             for(j=0;j<length2;j++){
 61                 if(number==array2[j])
 62                     break;
 63             }
 64 
 65             if(j==length2){
 66                 array2[length2]=number;
 67                 length2++;
 68             }
 69         }
 70 
 71         for(i=0;i<length2-1;i++){
 72             for(j=i+1;j<length2;j++){
 73                 if(array2[i]>array2[j]){
 74                     temp=array2[i];
 75                     array2[i]=array2[j];
 76                     array2[j]=temp;
 77                 }
 78             }
 79         }
 80 
 81         /*for(i=0;i<length2;i++)
 82             printf("%d ",array2[i]);
 83         printf("\n");*/
 84 
 85         if(length1!=length2){
 86             printf("NO\n");
 87             continue;
 88         }
 89 
 90         for(i=0;i<length1;i++){
 91             if(array1[i]!=array2[i]){
 92                 printf("NO\n");
 93                 break;
 94             }
 95         }
 96 
 97         if(i==length1)
 98             printf("YES\n");
 99 
100     }
101         
102     return 0;
103 }

 

posted @ 2014-11-06 21:36  zqxLonely  阅读(296)  评论(0编辑  收藏  举报