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Friday, September 6, 2013
SPOJ:Solution to the Fear of Three for Ross problem (simply lovable)
Question :
You know what? Ross had three marriages. All unsuccessful :D Because of this, he is angry with everyone. Whenever he gets a list of numbers during his research work, he wants to remove the maximum 3 numbers out of it to show his anger. He asked you to write a program for him, that takes a list of numbers and prints the 3 maximum numbers in it in non increasing order.
You know what? Ross had three marriages. All unsuccessful :D Because of this, he is angry with everyone. Whenever he gets a list of numbers during his research work, he wants to remove the maximum 3 numbers out of it to show his anger. He asked you to write a program for him, that takes a list of numbers and prints the 3 maximum numbers in it in non increasing order.
Inputs
First Line will contain "N" the no. of elements in the list. Next Line will contain "N" integers.
First Line will contain "N" the no. of elements in the list. Next Line will contain "N" integers.
Outputs :
The only line of output should contain 3 integers, that are maximum in the list in non increasing order.
The only line of output should contain 3 integers, that are maximum in the list in non increasing order.
Constraints
3 <= N <= 100, Numbers in the list can be between 1 and 1000 inclusive.
3 <= N <= 100, Numbers in the list can be between 1 and 1000 inclusive.
Level :Starter,Cakewalk
Solution to the Problem :
#include<stdio.h>
void perform(int *arr,int n)
{
int max1,max2,max3;
max1=arr[0];
int i,j,k;
int l;
i=0;
for(l=1;l<n;l++)
{
if(arr[l]>max1)
{
i=l;
max1=arr[i];
}
}
if(i==0)
{
j=1;
max2=arr[1];
}
else
{
j=0;
max2=arr[0];
}
for(l=0;l<n;l++)
{
if(l==i)
continue;
if(arr[l]>max2)
{
j=l;
max2=arr[j];
}
}
max3=-1;
for(l=0;l<n;l++)
{
if(l==i||l==j)
continue;
if(arr[l]>max3)
{
k=l;
max3=arr[k];
}
}
printf("%d %d %d\n",max1,max2,max3);
}
main()
{
int n,i;
scanf("%d",&n);
int arr[n];
for(i=0;i<n;i++)
scanf("%d",&arr[i]);
perform(arr,n);
}
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