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Thought of the Day

 Thought of the Day 

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  • We only see that, which our mind is ready to comprehend.  

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Find Remainder | FLOW002

Problem Write a program to find the remainder when an integer  A  is divided by an integer  B . Input The first line contains an integer  T , the total number of test cases. Then  T  lines follow, each line contains two Integers  A  and  B . Output For each test case, find the remainder when  A  is divided by  B , and display it in a new line. Constraints 1  ≤   T   ≤  1000 1  ≤   A,B   ≤  10000 Sample 1: Input Output 3 1 2 100 200 40 15 1 100 10 Solution: #include<iostream> using namespace std; int main(){     int t;     cin>>t;     while(t--){         int a, b;         cin>>a>>b;         cout<<a%b<<endl;     }     return 0; }
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Id and Ship | FLOW010

Problem Write a program that takes in a letterclass ID of a ship and display the equivalent string class description of the given ID. Use the table below. Class ID Ship Class B or b BattleShip C or c Cruiser D or d Destroyer F or f Frigate Input Format The first line contains an integer  T , the total number of testcases. Then  T  lines follow, each line contains a character. Output Format For each test case, display the Ship Class depending on ID, in a new line. Constraints 1  ≤   T   ≤  1000 Sample 1: Input Output 3 B c D BattleShip Cruiser  Destroyer   Solution: #include <iostream> using namespace std; int main() { // your code goes here int t; char ch; cin>>t; while(t--) { cin>>ch; if(ch=='B' || ch=='b') cout<<"BattleShip"<<endl; if(ch=='C' || ch=='c') cout<<"Cruiser"<<endl; if(ch=='D' || ch=='d') cout<<"Destroyer"<...
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Sum of Digits | FLOW006

  Problem You're given an integer  N . Write a program to calculate the sum of all the digits of  N . Input Format The first line contains an integer  T , the total number of testcases. Then follow  T  lines, each line contains an integer  N . Output Format For each test case, calculate the sum of digits of  N , and display it in a new line. Constraints 1 ≤ � ≤ 1000 1 ≤ T ≤ 1000 1 ≤ � ≤ 1000000 1 ≤ N ≤ 1000000 Sample 1: Input Output 3 12345 31203 2123 15 9 8 Solution: #include <bits/stdc++.h> using namespace std; int main() {     int t;     cin>>t;     while(t--)     {         int n;         cin>>n;         int sum = 0;         while(n!=0)         {             int r = n%10;             sum+= r;           ...
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