2019年杭电多校第三场 1008题Game(HDU6610+带修改莫队+Nim博弈)

题目链接

传送门

题意

给你\(n\)堆石子，每堆有\(a_i\)堆石子，\(q\)次操作：

• 在\([L,R]\)内有多少个子区间使得\(Alice\)(先手)在\(Nim\)博弈中获胜；
• 交换\(a_{pos},a_{pos+1}\)的值。

思路

这题和cf617E差不多。

首先我们知道以下性质：

• \(Nim\)博弈只有当所有石子数异或为\(0\)才会导致先手必败；
• 在预处理前缀异或和后，交换相邻两堆石子的石子数只会影响\(pos\)处的值。

因此我们在预处理出前缀异或和后就可以用待修改莫队来解决本题，我们用\(cnt\)数组来处理出区间内\(x\)出现次数，\(sum\)表示区间内有多少个子区间异或和为\(0\)，那么最后答案为\(\frac{(R-L+1)(R-L)}{2}-sum\)。

代码实现如下

#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cassert>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <unordered_map>
using namespace std;

typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;

#define lson rt<<1
#define rson rt<<1|1
#define lowbit(x) x&(-x)
#define name2str(name) (#name)
#define bug printf("*********\n")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
#define FIN freopen("D://Code//in.txt","r",stdin)
#define IO ios::sync_with_stdio(false),cin.tie(0)

const double eps = 1e-8;
const int mod = 1000000007;
const int maxn = 1e5 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3fLL;

int n, q, op, l, r, block;
LL sum;
int a[maxn], b[maxn];
LL cnt[1<<20+5];

struct que {
    int l, r, id, t;
    LL ans;
    bool operator < (const que& x) const {
        if((l - 1) / block != (x.l - 1) / block) {
            return l < x.l;
        }
        if((r - 1) / block != (x.r - 1) / block) {
            return r < x.r;
        }
        return t < x.t;
    }
}ask[maxn];

struct modify {
    int pre, val, pos;
}mfy[maxn];

void del(int x) {
    --cnt[x];
    sum -= cnt[x];
}

void add(int x) {
    sum += cnt[x];
    ++cnt[x];
}

int main() {
#ifndef ONLINE_JUDGE
FIN;
#endif // ONLINE_JUDGE
    while(~scanf("%d%d", &n, &q)) {
        sum = 0;
        for(int i = 0; i <= 1024 * 1024; ++i) cnt[i] = 0;
        for(int i = 1; i <= n; ++i) {
            scanf("%d", &a[i]);
            b[i] = b[i-1] ^ a[i];
        }
        block = (int)pow(n, 2.0 / 3);
        int nw = 0, idq = 0, idc = 0;
        for(int i = 1; i <= q; ++i) {
            scanf("%d", &op);
            if(op == 1) {
                idq++;
                scanf("%d%d", &ask[idq].l, &ask[idq].r);
                ask[idq].id = idq;
                --ask[idq].l;
                ask[idq].t = nw;
            } else {
                idc++;
                ++nw;
                scanf("%d", &mfy[idc].pos);
                int pos = mfy[idc].pos;
                mfy[idc].pre = b[pos];
                mfy[idc].val = b[pos+1] ^ a[pos];
                b[pos] = b[pos+1] ^ a[pos];
                swap(a[pos], a[pos+1]);
            }
        }
        sort(ask + 1, ask + idq + 1);
        int tmp = nw;
        for(int i = 1, l = 1, r = 0; i <= idq; ++i) {
            while (r < ask[i].r) add(b[++r]);
            while (l > ask[i].l) add(b[--l]);
            while (r > ask[i].r) del(b[r--]);
            while (l < ask[i].l) del(b[l++]);
            while (tmp < ask[i].t) {
                tmp++;
                if (mfy[tmp].pos >= ask[i].l && mfy[tmp].pos <= ask[i].r) {
                    del(mfy[tmp].pre);
                    add(mfy[tmp].val);
                }
                b[mfy[tmp].pos] = mfy[tmp].val;
            }
            while (tmp > ask[i].t) {
                if (mfy[tmp].pos >= ask[i].l && mfy[tmp].pos <= ask[i].r) {
                    del(mfy[tmp].val);
                    add(mfy[tmp].pre);
                }
                b[mfy[tmp].pos] = mfy[tmp].pre;
                tmp--;
            }
            ask[ask[i].id].ans = 1LL * (ask[i].r - ask[i].l) * (ask[i].r - ask[i].l + 1) / 2 - sum;
        }
        for(int i = 1; i <= idq; ++i) {
            printf("%lld\n", ask[i].ans);
        }
    }
    return 0;
}