hdu 5565 Clarke and baton（水）

题目链接：hdu 5565 Clarke and baton

代码

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>

using namespace std;
typedef long long ll;
const int maxn = 1e7 + 1;
const int mod = 1e9 + 7;
const int G = 78125;

int N, Q, S, T, A[maxn];
deque<int> L[maxn], R[maxn];

int mrand(int l, int r) {
    S = 1LL * S * G % mod;
    return l + S % (r-l+1);
}

void init () {
    scanf("%d%d%d", &N, &Q, &S);
    T = 0;
    int sum = mrand(Q, 10000000);
    for(int i = 1; i <= N; i++) {
        A[i] = mrand(0, sum/(N-i+1));
        sum -= A[i];
        T = max(T, A[i]);
    }
    int tmp = mrand(1, N);
    A[tmp] += sum;
    T = max(T, A[tmp]);

    for (int i = 0; i <= T; i++) {
        L[i].clear();
        R[i].clear();
    }
    for (int i = 1; i <= N; i++)
        L[A[i]].push_back(i);
}

int solve () {
    while (Q--) {
        if (L[T].empty() && R[T].empty()) T--;

        if (R[T].empty() || (!L[T].empty() && L[T].front() < R[T].front())) {
            R[T-1].push_back(L[T].front());
            L[T].pop_front();
        } else {
            R[T-1].push_back(R[T].front());
            R[T].pop_front();
        }
    }

    for (int i = 0; i < T; i++) {
        for (int j = 0; j < L[i].size(); j++) A[L[i][j]] = i;
        for (int j = 0; j < R[i].size(); j++) A[R[i][j]] = i;
    }

    int ans = 0;
    for (int i = 1; i <= N; i++)
        ans ^= (A[i] + i);
    return ans;
}

int main () {
    int cas;
    scanf("%d", &cas);
    while (cas--) {
        init();
        printf("%d\n", solve());
    }
    return 0;
}