Uva 11991 - Easy Problem from Rujia Liu?

By -
/*

The Problem is vary impressive. Find the number's index among such queries. But that does not matter. The matter is Time limit 1sec.
If you do linear search than your complexity may be O(N^2) or more, so, that will be inefficient for that time.
Another Process, You may do Binary Search which has complexity O(lgN), but we know that the precondition of binary search is
to Sort data first. That means it will take O(NlgN) complexity. So, it will also inefficient for that time.
I have a method to solve that problem by using Precalculation method or DP.

Uva 11991 - Easy Problem from Rujia Liu?
Verdict : Accepted
Time: 0.900
Solution Method: Structure, DP

*/

#include<bits/stdc++.h>
#define N 1000010
#define sf scanf
#define pf printf
using namespace std;

typedef long long LL;
typedef map<long long, long long> mpll;
typedef map<long long, bool>mpbll;

struct my
{
    LL dx, num, cnt;
};

my starr[N];
LL numarr[N], parr[N], plen=0 , starrlen=1;

void input_n(LL k)
{
    mpll mp1;
    mp1.clear();
    LL mpocnt=0;

    for(LL i=1;i<=k;i++)
    {
        LL x;
        sf("%lld",&x);
        mp1[x]++; // Counting map. I have just count how many times a number exist. complexity O(lgN)
        mpocnt = mp1[x];
        numarr[i] = x;
        starr[x].num=x;
        starr[x].dx = i;
        starr[x].cnt=mpocnt;
    }

//    for(LL i=1;i<=starrlen; i++)
//    {
//        cout << starr[i].num << " " << starr[i].cnt << " " << starr[i].dx << "\n";
//    }
}

int main()
{
    LL n,m;

    while(~sf("%lld %lld",&n,&m))
    {
        plen=0;

        input_n(n);

        while(m--)
        {
            LL kotobar, nm;
            sf("%lld %lld",&kotobar,&nm);

            if(starr[nm].cnt == kotobar)
            {
                parr[plen++]=starr[nm].dx;
            }

            else if(starr[nm].cnt > kotobar)
            {
                LL i, chkcnt=0;
                bool f=false;

                for(i=1; i<=starr[nm].dx; i++)
                {
                    if(nm == numarr[i])
                    {
                        chkcnt++;
                    }

                    if(chkcnt == kotobar)
                    {
                        parr[plen++]=i;
                        break;
                    }
                }
            }

            else
            {
                parr[plen++]=0;
            }
        }

        for(LL i=0;i<plen;i++)
        {
            pf("%lld\n",parr[i]);
        }
    }

    return 0;
}

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