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  1. #include <stdio.h>
  2. int find_min(int a,int b){
  3. 	if (a > b){
  4. 		return b;}
  5. 	else  {
  6. 		return a;}}
  7.  
  8. int find_intersection_with_count(int *arr1,int *arr2,int n1,int n2){
  9. 	int count1[10] = {0};
  10. 	int count2[10] = {0};
  11. 	int count[10] = {0};
  12. 	for (int i = 0;i <= 9;i ++){
  13. 		for (int j = 0;j < n1;j ++){
  14. 			if(*(arr1 + j) == i){
  15. 				count1[i] ++;}}
  16. 		for (int j = 0;j < n2;j ++){
  17. 			if(*(arr2 + j) == i){
  18. 				count2[i] ++;}}
  19. 		count[i] = find_min(count1[i],count2[i]);
  20. 		if (count[i] != 0){
  21. 			printf("%d (%d) ",i,count[i]);}}
  22. 	printf("\n");
  23. 	for (int k = 9;k >= 0 ;k --){
  24. 		if (count[k] != 0){
  25. 			printf("%d (%d) ",k,count[k]);}}}
  26.  
  27. int main(void) {
  28. 	int n1,n2;
  29. 	scanf("%d",&n1);
  30. 	int arr1[n1];
  31. 	for (int i = 0;i < n1;i ++){
  32. 		scanf("%1d",&arr1[i]);}
  33. 	scanf("%d",&n2);
  34. 	int arr2[n2];
  35. 	for (int i = 0;i < n2;i ++){
  36. 		scanf("%1d",&arr2[i]);}
  37. 	find_intersection_with_count(&arr1,&arr2,n1,n2);
  38. 	return 0;}
Success #stdin #stdout 0.01s 5272KB
comments (?)
stdin
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10
5 7 8 8 1 8 5 8 7 8
6
5 7 8 8 8 8
stdout
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5 (1) 7 (1) 8 (4) 
8 (4) 7 (1) 5 (1)
https://ideone.com/GVfh3P
assiment3
language:
C (gcc 8.3)
created:
15 hours ago
visibility:
public

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