LightOj 1354 - IP Checking

By -
// Verdict:: Accepted
  1. #include <iostream>
  2. #include <cstdio>
  3. #include <cstring>
  4. #include <string>
  5. #include <vector>
  6. #include <cmath>
  7. #include <cctype>
  8. #include <set>
  9. #include <list>
  10. #include <stack>
  11. #include <queue>
  12. #include <algorithm>
  13. #define sf scanf
  14. #define pf printf
  15. #define L long
  16. #define LL long long
  17. #define N 10000
  18. #define nl puts("")
  19. #define MOD 10000000007
  20. #define pb push_back
  21. #define ps push
  22. #define clr clear
  23. #define bn begin()
  24. #define ed end()
  25.  
  26. using namespace std;
  27.  
  28. int main()
  29. {
  30.     int tc,t;
  31.     char ch1[10],ch2[10],ch3[10],ch4[10];
  32.     sf("%d",&tc);
  33.     for (t=1;t<=tc;t++)
  34.     {
  35.         int a,b,c,d;
  36.         LL b1,b2,b3,b4;
  37.         sf("%d.%d.%d.%d",&a,&b,&c,&d);
  38.         sf("%lld.%lld.%lld.%lld",&b1,&b2,&b3,&b4);
  39.         int s1=0,s2=0,s3=0,s4=0,p1,p2,p3,p4,len1,len2,len3,len4;
  40.         sprintf(ch1,"%lld",b1);
  41.         len1=strlen(ch1);
  42.         p1=len1-1;
  43.  
  44.         sprintf(ch2,"%lld",b2);
  45.         len2=strlen(ch2);
  46.         p2=len2-1;
  47.  
  48.         sprintf(ch3,"%lld",b3);
  49.         len3=strlen(ch3);
  50.         p3=len3-1;
  51.  
  52.         sprintf(ch4,"%lld",b4);
  53.         len4=strlen(ch4);
  54.         p4=len4-1;
  55.  
  56.         for (int i=0;i<len1;i++)
  57.         {
  58.             s1+=(ch1[i]-48)*pow(2,p1);
  59.             p1--;
  60.         }
  61.  
  62.         for (int i=0;i<len2;i++)
  63.         {
  64.             s2+=(ch2[i]-48)*pow(2,p2);
  65.             p2--;
  66.         }
  67.  
  68.         for (int i=0;i<len3;i++)
  69.         {
  70.             s3+=(ch3[i]-48)*pow(2,p3);
  71.             p3--;
  72.         }
  73.  
  74.         for (int i=0;i<len4;i++)
  75.         {
  76.             s4+=(ch4[i]-48)*pow(2,p4);
  77.             p4--;
  78.         }
  79.  
  80.         if(s1==a and s2==b and s3==c and s4==d)
  81.         {
  82.             pf("Case %d: Yes\n",t);
  83.         }
  84.         else
  85.         {
  86.             pf("Case %d: No\n",t);
  87.         }
  88.     }
  89.  
  90.     return 0;
  91. }

Comments

Post a Comment

Popular posts from this blog

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";         pf("Case %d:\n",++tc);         getchar();         while(n--)         {             //cin >> c;             sf("%c",&c);             if(c == 'A')             {                 //cin >> x;           
Read more

LightOJ 1009 - Back to Underworld

By -
// Accepted /* Analysis: Consider a Graph as 1-2, 2-3, 2-4; if 1 is a vampire than 2 is lykan, as 2 is lykan so 3 and 4 will be vampire. Now total vampires are 3 and lykan is 1, so maximum from this graph is 3. [1=3=4=vampire] another graph is 5-6 and 6-7, 6-8. consider 5 is a vampire so 6 will be lykan. as 6 is lykan than the all adjacent of 6 will be vampire, i mean, 7=8=vampire. Now total vampires are 3 and lykan is 1, so maximum from this graph is 3. And the Answer of this total Graph is 3 + 3 = 6; [As, Graphs are not connected] Decision: If a node is Vampire than all of the adjacent of that node will be Lykan and if a node is Lykan than all of the adjacent of that node will be Vampire, thats why i have represented them as two color black and red. Black means Vampire and Red means Lykan, besides every time i have increased number of vampire and lykans and finally add the maximum as graphs are not connected. */ #include<cstdio> #include<cmath> #include<vector> #in
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 )         {             digit [len++] = x% 10 ;             x/= 10 ;         }         for ( int i= 0 , j=len- 1 ; i<len; i++, j--)         {             if ( digit [i] != digit [j])             {  
Read more