Skip to main content

BADXOR

BADXOR - Bad XOR


#include <iostream>
#include <cstring>

#define REP(i, n) for(int i = 0; i < n; i++)
#define REP1(i, n) for(int i = 1; i <= n; i++)

using namespace std;

typedef long long int ll;

const int maxLimit = 100000007;

ll dp[1005][1024];
//dp[i][j] = number of ways of making value j with first i elements

int main()
{
    int t, a, b;
    int arr1[1001];
    bool arr2[1025];
    cin >> t;

    REP1(k, t)
    {
        memset(dp, 0, sizeof(dp));
        memset(arr2, false, sizeof(arr2));

        cin >> a >> b;

        REP(i, a)
        {
            cin >> arr1[i];
        }

        REP(i, b)
        {
            int temp;
            cin >> temp;
            arr2[temp] = true;
        }

        dp[0][0] = 1;

        REP1(i, a)
        {
            REP(j, 1024)
            {
                dp[i][j] = dp[i - 1][j] + dp[i - 1][j ^ arr1[i - 1]];

                if(dp[i][j] >= maxLimit)
                    dp[i][j] -= maxLimit;
            }
        }


        ll total = 0; // number of subset

        REP(i, 1024)
        {
            if(!arr2[i])
            {
                total += dp[a][i];
                if(total >= maxLimit)
                total -= maxLimit;
            }

        }

        cout << "Case " << k << ": " << total << endl;

    }
    return 0;
}

Comments

  1. Unknown21 December 2018 at 23:34

    can somesone post the solution of BADXOR using topdown approach

    ReplyDelete

Post a Comment

Popular posts from this blog

GCJ101BB - Picking Up Chicks

Problem Link /* explanation     lets solve the problem only for 2 chicken.     s[i] = speed of chicken i     pos[i] = position of chicken i     if s[i] > s[i - 1] then no problem, just check whether both can reach b within time or not.     if s[i] < s[i - 1] then there is a chance that i can slow down i - 1.     lets say s[i] = 1 m/sec and s[i - 1] = 2 m/sec and time limit is T and point to reach is B.     for s[i] pos[i] can be at max B - T. if pos is greater than B-T it can not reach within Tsec.     and for s[i - 1] pos[i - 1] can be at max (B-T)*2. if pos[i - 1] > (B-T)*2 it can not reach within Tsec.     at T sec i will be at B and i - 1 will also be at B. at T - 1 i will be at B-T-1 and i-1 will be at B-T-2 and so on. as we can see i -1 will always be behind i. so there will not be any collision.     if i is pos[i] < B-T then i can reach B before T sec ...
Post a Comment
Read more

KOPC12A

KOPC12A - K12 - Building Construction #include <iostream> #include <cmath> #define REP(i, n) for(int i = 0; i < n; i++) using namespace std; const int N = 10005; int height[N]; int cost[N]; int n; //finds the total cost for height h long long int findCost(int h) {     long long c = 0; //cost     REP(i, n)     {         c += abs(h - height[i]) * cost[i];     }     return c; } int ternary_search(int l, int h) {     while(l <= h)     {         if(l == h)             break;         int mid1 = l + (h - l) / 3;         int mid2 = h - (h - l) / 3;         if(findCost(mid1) > findCost(mid2))             l = mid1 + 1;         else ...
Post a Comment
Read more

Yet Another Cute Girl - PRETNUM

Problem Link #include <iostream> #include <cstring> #include <cmath> #include <vector> using namespace std; #define LL long long const int N = 1000010; bool prime[N + 1]; vector<int> v; void sieve() { for(int i = 2; i <= N; i++) { prime[i] = true; } for(int i = 2; i * i <= N; i++) { if(prime[i]) { for(int j = i * i; j <= N; j += i) { prime[j] = false; } } } v.push_back(2); for(int i = 3; i <= N; i += 2) { if(prime[i]) { v.push_back(i); } } } LL getNumDiv(LL num) {     /* if prime factor of n is p1^k1 p2 ^ k2 then prime factor of n^2 would be         (p1^k1 p2 ^ k2)^2 = p1^(2k1) p2^(2k2)         so number of divisors of n^2 would be (2 * k1 + 1) * (2 * k2 + 1) ...         if n is a prime number number then number of divisors of n^2 would be 3.     */ LL res = 1; for(int i = 0;...
Post a Comment
Read more