杭电多校（二）2019.7.24--暑假集训

【HDU 6591】

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概率期望，算是一道纯数学题

#include<cstdio>
#include<iostream>
#define ll long long
const int MAXN = 3001;
const int mod  = 998244353;
ll ans[MAXN];
ll qpower(ll a, ll x)
{
    ll res = 1;
    ll down = a;
    while (x)
    {
        if (x & 1)
        {
            res = (res*down) % mod;
        }
        down = (down*down) % mod;
        x = x >> 1;
    }
    return res;
}
void init()
{
    for (int i = 1; i <MAXN; i++)
    {
        ans[i] = ((i * (i - 1))%mod)*qpower(3, mod - 2);
        ans[i] %= mod;
        ans[i] = (ans[i] + ans[i - 1]) % mod;
    }
    for (int i = 1; i < MAXN; i++)
    {
        ans[i] = (ans[i] * qpower(i, mod - 2))%mod;
    }
}
signed main()
{
    int n;
    init();
    while (~scanf("%d",&n))
    {
        printf("%lld\n", ans[n]);
    }
    return 0;
}

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不平等博弈

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建图+网络流

回文

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【题目大意】

给定一个长度为n的整数序列，给定m次询问，每次询问区间内整数能够合法组成三角形的最大长度

【思路】

基本上是裸的主席树，找到区间第1大，第2大，第3大，然后计算合法性，如果不合法，找第2大，第3大，第4大，以此类推

因为三角形边长的合法情况是一个斐波那契数列，所以查找次数不会超过44次，时间复杂度较小

【知识点】：主席树

#include<cstdio>
#include<iostream>
#include<algorithm>
#define ll long long
using namespace std;
const int MAXN = 200010;
const int MAXM = 200010;
int lsan[MAXN], a[MAXN],T[MAXN];
int sum[21*MAXN], ls[21 * MAXN], rs[21 * MAXN];
int n, m;
int cnt;
int build(int l, int r)
{
    int root=++cnt;
    sum[root] = 0;
    int mid = (l + r) >> 1;
    if (l < r)
    {
        ls[root] = build(l, mid);
        rs[root] = build(mid + 1, r);
    }
    return root;
}
int update(int pre, int l, int r,int x)
{
    int root=++cnt;
    sum[root] = sum[pre] + 1;
    ls[root] = ls[pre];
    rs[root] = rs[pre];
    int mid = (l + r) >> 1;
    if (l < r)
    {
        if (x <= mid)
            ls[root] = update(ls[pre], l, mid, x);
        else
            rs[root] = update(rs[pre], mid + 1, r,x);
    }
    return root;
}
int query(int l,int r,int x, int y, int num)
{
    if (l >= r)
        return l;
    int t = sum[rs[y]] - sum[rs[x]];
    int mid = (l + r) >> 1;
    if (t>=num)
        return query(mid + 1, r, rs[x], rs[y], num);
    else
        return query(l, mid, ls[x], ls[y], num-t);
}
int main()
{
    while (~scanf("%d%d", &n, &m))
    {
        cnt = 0;
        for (int i = 1; i <= n; i++)
        {
            scanf("%d", &a[i]);
            lsan[i] = a[i];
        }
        sort(lsan + 1, lsan + 1 + n);
        int maxn = unique(lsan + 1, lsan + 1 + n) - lsan - 1;
        T[0] = build(1, maxn);
        for (int i = 1; i <= n; i++)
        {
            int t = lower_bound(lsan + 1, lsan + 1 + maxn, a[i]) - lsan;
            T[i] = update(T[i - 1], 1, maxn, t);
        }
        for (int i = 1; i <= m; i++)
        {
            int x, y, k;
            scanf("%d%d", &x, &y);
            int cal = 0;
            ll l1=0, l2=0, l3=0;
            cal = 0;
            while ((l1 + l2 <= l3 || l1 + l3 <= l2 ||l2 + l3 <= l1) &&cal+3<=y+1-x)
            {
                l1 = lsan[query(1, maxn, T[x - 1], T[y], ++cal)];
                l2 = lsan[query(1, maxn, T[x - 1], T[y], ++cal)];
                l3 = lsan[query(1, maxn, T[x - 1], T[y], ++cal)];
                cal -= 2;
            }
            if (l1 + l2 > l3 && l1 + l3 > l2 &&l2 + l3 > l1)
                printf("%lld\n", l1 + l2 + l3);
            else
                printf("-1\n");
        }
    }
    return 0;
}

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线段树

原文地址：https://www.cnblogs.com/rentu/p/11241030.html