目录

• 题目链接
• 题意
• 思路
• 代码

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;
}

标签：Stones,typedef,lb,Removing,石子,mid,pos,include,第三场
来源： https://www.cnblogs.com/Dillonh/p/11391423.html