codeforces 285(树，康托展开，组合，二分)

501C Misha and Forest

只要看出这是一个包含若干棵树的森林就好办了，从叶子开始删除。。

504B - Misha and Permutations Summation
Factorial number system 或者

康托展开

康托展开 是 排列 和 十进制整数之间的双向映射，其值刚好就是排列的次序。

但是题目数据范围没办法直接做，但转换的中间结果可以表示在中 Factorial number system  中。。

因为在转换的时候，需要快速求 k是集合中第几大， 集合中第k大的数， 以及要从集合中删除数，所以可以用 BIT 或者 Treap

#include<bits/stdc++.h>
using namespace std;

#define SPEED_UP iostream::sync_with_stdio(false);
#define FIXED_FLOAT cout.setf(ios::fixed, ios::floatfield);
#define rep(i, s, t) for(int (i)=(s);(i)<=(t);++(i))
#define urep(i, s, t) for(int (i)=(s);(i)>=(t);--(i))

typedef long long LL;

const int Maxn = 200000 + 100;

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

int c[Maxn], p[Maxn], q[Maxn], r[Maxn];
int n;

int sum(int x) {
    int ret = 0;
    while (x > 0) {
        ret += c[x];
        x -= lowbit(x);
    }
    return ret;
}

void update(int x, int v) {
    while (x <= n) {
        c[x] += v;
        x += lowbit(x);
    }
}

vector<int> Ord(int a[]) {
    memset(c, 0, sizeof(c));
    rep(i, 1, n) update(i, 1);

    vector<int> ret;
    rep(i, 0, n-1) {
        ret.push_back(sum(a[i]+1) - 1);
        update(a[i]+1, -1);
    }

    return ret;
}

int b_find(int k) {
    int l = 1, r = n, mid;
    while (l < r) {
        mid = (r+l)/2;
        int tmp = sum(mid);
        if (tmp < k)
            l = mid+1;
        else
            r = mid;
    }
    return l;
}

vector<int> Perm(int a[]) {
    memset(c, 0, sizeof(c));
    rep(i, 1, n) update(i, 1);

    vector<int> ret;
    rep(i, 0, n-1) {
        //cout << "i: " << i << " find " << a[i]
        int tmp = b_find(a[i]+1);
        update(tmp, -1);
        ret.push_back(tmp-1);
    }

    return ret;
}

void print(vector<int> & v) {
    for (auto i:v) cout << i << ' ';cout << endl;
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("input.in", "r", stdin);
#endif
    SPEED_UP
    cin >> n;
    rep(i, 0, n-1) cin >> p[i];
    rep(i, 0, n-1) cin >> q[i];

    vector<int> ord1 = Ord(p);
    vector<int> ord2 = Ord(q);

    //cout << "p -> ";print(ord1);
    //cout << "q -> ";print(ord2);

    int z = 0;
    urep(i, n-1, 0) {
        z += ord1[i] + ord2[i];
        r[i] = z % (n - i);
        z /= (n-i);
    }

    //rep(i, 0, n-2) cout << r[i] << ' ';cout << r[n-1] << endl;

    vector<int> ans = Perm(r);
    rep(i, 0, n-2) cout << ans[i] << ' ';cout << ans[n-1] << endl;

    return 0;
}

504C - Misha and Palindrome Degree 

观察 = =

序列下标： 0 - n-1

首先，去掉首尾相等的部分，即 存在k， 所有 i < k,  a[i] == a[n-i-1] , a[k] != a[n-k-1],  L = k, R = n-1-k

所以，所求的 range 必然包含 L 或者 R

于是问题转换为求出 [L, R] 的最小前缀range 和最小后缀range，

++..+.L.....R+...++   

可以用二分和O(n)方法

O(n) 方法是这样的， 可以再观察一下。。不论是前缀range还是后缀range，都存在一个上限，即[L, R]里的每一种元素，都有一半在range里面。可以先求出这个上限，然后缩减。

最后总的求和的时候，应用容斥原理就可以方便的求出来了 = =

#include<bits/stdc++.h>
using namespace std;

#define SPEED_UP iostream::sync_with_stdio(false);
#define FIXED_FLOAT cout.setf(ios::fixed, ios::floatfield);
#define rep(i, s, t) for(int (i)=(s);(i)<=(t);++(i))
#define urep(i, s, t) for(int (i)=(s);(i)>=(t);--(i))

typedef long long LL;

const int Maxn = ((int)1e5) + 100;

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

int arr[Maxn], n;
map<int, int> num, numL, numR;

LL Solve(int a[]) {
    int l = 0, r = n-1;
    while (a[l] == a[r] && l < r) ++l, --r;
    if (l >= r) return n * 1ll * (n+1)/2;

    rep(i, l, r) num[a[i]] += 1;
    int odds = 0, key = -1;
    for (auto it = num.begin();it != num.end();++it)
        if ( (it->second)&1 )
            odds += 1, key = it->first;
    if (odds > 1) return 0;

    // 找最小前缀
    int oks = 0, l2 = l-1;
    while (oks < num.size()) {
        numL[a[++l2]] += 1;
        int _a = numL[a[l2]] << 1;
        int _b = _a - 2, _c = num[a[l2]];
        if ( _a >= _c && _b < _c ) ++oks;
    }

    oks = 0;
    int r2 = r+1;
    while (oks < num.size()) {
        numR[a[--r2]] += 1;
        int _a = numR[a[r2]] << 1;
        int _b = _a - 2, _c = num[a[r2]];
        if ( _a >= _c && _b < _c ) ++oks;
    }
    //cout << "be min prefix: " << l << ", " << l2 << "  min suffix: " << r2 << ", " << r << endl;
    int mid = (r+l)>>1;
    if (l2 <= mid && ( !(n&1) || a[mid] == key )) {
        int g, h;
        if (n&1)
            g = mid-1, h = mid+1;
        else
            g = mid, h = mid+1;
        while (a[g] == a[h]) --g, ++h;
        l2 = g, r2 = h;
    }
    //cout << "min prefix: " << l << ", " << l2 << "  min suffix: " << r2 << ", " << r << endl;
    LL sum = 0;
    sum += (l+1) * 1ll * (n-l2);
    sum += (n-r) * 1ll * (r2+1);
    sum -= (l+1) * 1ll * (n-r);

    return sum;
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("input.in", "r", stdin);
#endif
    SPEED_UP
    cin >> n;
    rep(i, 0, n-1) cin >> arr[i];

    cout << Solve(arr) << endl;
    return 0;
}