Uva 10469 - To Carry or not to Carry

By -
/*
    Uva 10469 - To Carry or not to Carry
    Verdict :: Accepted
    Time:: 0.000
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
using namespace std;

int main ()
{
    unsigned int a,b;

    while (cin >> a >> b)
    {
        int c = a^b;

        cout << c << endl;
        //printf("%u\n",c);
    }

    return 0;
}

Comments

Post a Comment

Popular posts from this blog

Uva 10650 - Determinate Prime

By -
/*     Verdict : Accepted     Link: Determinate Prime Solution Link */ #include<bits/stdc++.h> #define limit 32009 #define sf scanf #define pf printf using namespace std; typedef long long LL; typedef vector<int>vi; int prm[limit], plen=1; bitset<limit> bs; void sieve() {     LL i,j;     bs.set();     bs[1] = bs[0] = 0;     for(i=2; i<limit; i++)     {         if(bs[i])         {             if(i != 2)             {                 prm[plen++] = i;             }          ...
Read more

SPOJ-CMG - Collecting Mango

By -
#include<iostream> #include<cstdio> #include<cstring> #include<vector> #include<queue> #include<stack> #include<map> #include<cmath> #include<algorithm> #define psb push_back #define ppb pop_back #define all(x) x.begin(),x.end() #define N 1000100 #define sf scanf #define pf printf using namespace std; typedef long long LL; typedef vector<LL>vll; LL ar[N] , num[N]; int main() {     int t , tc=0;     //cin >> t;     sf("%d",&t);     while(t--)     {         LL n, x, arlen=-1, nmlen=-1 , i , maxi=-1 , zar;         char c;         //cin >> n;         sf("%lld",&n);         //cout << "Case " << ++tc << ":\n";       ...
Read more

LeetCode Palindrome Number

By -
 Problem Description Link:  https://leetcode.com/problems/palindrome-number/ Accepted Solution /* The solution is easy, and I believe everyone knows the process. However, one problem is that x is a number. If it is a string, in that case, we can get the result after reversing it. Therefore, to solve the problem, we have to find the digits by modulating 10, store them in an array, and finally compare from both sides. If we find any discrepancy, we can tell it is not a palindrome. Note that any negative integer would never be a palindrome since it should contain a minus (-) sign. */ class Solution { public:     bool isPalindrome ( int x ) {         int digit [ 50 ], len= 0 ;         bool f = true ;         if (x < 0 ) {             return false ;         }         while (x> 0 )         {       ...
Read more