Problem solution in C programming.
#include <stdio.h> int main(void) { int n,a,b; scanf("%d",&n); while (n--) { scanf("%d%d",&a,&b); printf("%d\n",a-b); } return 0; }
Here is the above code first we read an integer number N that's the number of a test case. and after that, we run a loop till then N got negative. and in the loop, we scan each test case value and print the amount of water that we need to fill in the bucket.
Problem solution in C++ programming.
#include <iostream> using namespace std; int main() { // your code goes here int t; cin>>t; while(t--) { int k,x; cin>>k; cin>>x; cout<<k-x<<endl; } return 0; }
Here in the above C++ code first we read an integer t for the test cases. and then we run a while loop to read all the test cases and print the amount of extra water we need to fill the bucket.
Problem solution in Java programming.
import java.util.*; class FBC { public static void main(String args[]){ Scanner sc= new Scanner(System.in); int t=sc.nextInt(); while(t-- > 0){ int x=sc.nextInt(); int y=sc.nextInt(); System.out.println(x-y); } } }
Problem solution in Python programming.
t=int(input()) for _ in range(t): k,x=map(int,input().split()) y=k-x print(y)
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