#include<stdio.h> | |
#include<stdlib.h> | |
int main(){ | |
int i,j,*arr,len,total=99999,sum,num1,num2; | |
scanf("%d",&len); | |
//Dynamically allocating memory for array | |
arr = (int*) malloc(len * sizeof(int)); | |
for(i=0;i<len;i++){ | |
scanf("%d",&arr[i]); | |
} | |
for(i=0;i<len;i++){ | |
for(j=0;j<len;j++){ | |
if(i==j) | |
break; | |
if(abs(arr[i]+arr[j])<abs(total)){ | |
total = arr[i]+arr[j]; | |
num1=arr[i]; | |
num2=arr[j]; | |
} | |
} | |
} | |
printf("sum = %d\nNumbers are %d %d",total,num1,num2); | |
} |
Tuesday 26 September 2017
Given an array with both positive and negative numbers. Find two elements such that their sum is closest to zero
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