题目链接

传送门

题意

初始时有\(n\)堆石子，每堆石子的石子个数为\(a_i\)，然后进行游戏。

游戏规则为你可以选择任意两堆石子，然后从这两堆中移除一个石子，最后石子个数变为\(0\)则获胜否则失败。由于总石子个数可能为奇数，此时不可能获胜，因此加了个规则为如果石子个数为奇数，那么可以事先移除一个石子。

问你有多少个区间能让玩游戏的人获胜。

思路

经过模型转换后题意变为有多少个区间，区间内石子个数之和大于等于石子最大数的两倍。

启发式分治，大体处理方法和这题一样。

代码

#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>

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),L,mid

#define rson (rt<<1|1),mid + 1,R

#define lowbit(x) x&(-x)

#define name2str(name) (#name)

#define bug printf("*********\n")

#define debug(x) cout<<#x"=["<<x<<"]" <<endl

#define FIN freopen("/home/dillonh/CLionProjects/Dillonh/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 = 300000 + 2;

const double pi = acos(-1);

const int inf = 0x3f3f3f3f;

const LL INF = 0x3f3f3f3f3f3f3f3fLL;

LL ans;

int _, n;

LL sum[maxn];

int a[maxn], dp[maxn][20], pos[maxn][20], lg[maxn];

void init() {

    lg[0] = -1;

    for(int i = 1; i <= n; ++i) lg[i] = lg[i>>1] + 1;

    for(int j = 1; j <= lg[n]; ++j) {

        for(int i = 1; i + (1<<j) - 1 <= n; ++i) {

            if(dp[i][j-1] >= dp[i+(1<<(j-1))][j-1]) {

                dp[i][j] = dp[i][j-1];

                pos[i][j] = pos[i][j-1];

            } else {

                dp[i][j] = dp[i+(1<<(j-1))][j-1];

                pos[i][j] = pos[i+(1<<(j-1))][j-1];

            }

        }

    }

}

int query(int l, int r) {

    int k = lg[r-l+1];

    if(dp[l][k] >= dp[r-(1<<k)+1][k]) return pos[l][k];

    else return pos[r-(1<<k)+1][k];

}

void solve(int l, int r) {

    if(l >= r) return;

    if(l + 1 == r) {

        ans += (a[l] == a[r]);

        return;

    }

    int pos = query(l, r);

    if(r - pos > pos - l) {

        for(int i = l; i <= pos; ++i) {

            int ub = r, lb = pos + 1, mid, pp = -1;

            if(i != pos) lb = pos;

            while(ub >= lb) {

                mid = (ub + lb) >> 1;

                if(sum[mid] - sum[i-1] >= 2 * a[pos]) {

                    pp = mid;

                    ub = mid - 1;

                } else {

                    lb = mid + 1;

                }

            }

            if(pp == -1) continue;

            ans += (r - pp + 1);

        }

    } else {

        for(int i = pos; i <= r; ++i) {

            int ub = pos - 1, lb = l, mid, pp = -1;

            if(i != pos) ub = pos;

            while(ub >= lb) {

                mid = (ub + lb) >> 1;

                if(sum[i] - sum[mid-1] >= 2 * a[pos]) {

                    pp = mid;

                    lb = mid + 1;

                } else {

                    ub = mid - 1;

                }

            }

            if(pp == -1) continue;

            ans += (pp - l + 1);

        }

    }

    solve(l, pos - 1), solve(pos + 1, r);

}

int main() {

#ifndef ONLINE_JUDGE

    FIN;

#endif

    scanf("%d", &_);

    while(_--) {

        scanf("%d", &n);

        for(int i = 1; i <= n; ++i) {

            scanf("%d", &a[i]);

            sum[i] = sum[i-1] + a[i];

            dp[i][0] = a[i];

            pos[i][0] = i;

        }

        init();

        ans = 0;

        solve(1, n);

        printf("%lld\n", ans);

    }

    return 0;

}

巴特西

Removing Stones(2019年牛客多校第三场G+启发式分治)

题目链接

题意

思路

代码

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