COCI2017-2018#3 Dojave || 洛谷P4443

题目传送门.....................................................................................................................................

The biggest event of the year ended tragically for Croatian teams. The most influential theoretician of CERC of all time, the founder of the popular page CERC Tips, and in his free time an outstanding bass player, in his most recent performance failed to get his team to the finals.

In order to get over his existential troubles, our subject is spending time playing games of chance. He is especially interested in the following game:

You are given a positive integer M. Our protagonist sees in front of him a permutation of an array of numbers 0, 1, 2, ..., 2^M2M - 1.

The computer chooses a nonempty contiguous subsequence of the given permutation, which it then lights up over a capital city of one of the countries in Southeastern Europe.

Our confidant, after fighting off tears caused by memories of old times, must choose two distinct elements of the permutation and swap their places. Our man of the hour wins if and only if the bitwise XOR of the numbers in the lit up subsequence after the substitution is precisely 2^M2M - 1.

Our hero wants to know the number of contiguous subsequences the computer can light up so that he can win.

Help our hero overcome his (id)entity crisis so our favourite page can be fully active again.

输入输出格式

输入格式：

The first line of input contains the integer M (1 ≤ M ≤ 20),

The following line contains 2^M2M space-separated numbers that make up a permutation of the array 0, 1, 2, ..., 2^M2M - 1.

输出格式：

You must output the total number of contiguous subsequences that a computer can light up so our hero can win.

输入输出样例

输入样例#1：复制

2
0 1 2 3

输出样例#1：复制

输入样例#2：复制

3
3 7 0 4 6 1 5 2

输出样例#2：复制

输入样例#3：复制

4
13 0 15 12 4 8 7 3
11 14 6 10 1 5 9 2

输出样例#3：复制

说明

In test cases worth 50% of total points, it will hold 1 ≤ M ≤ 14.

Clarification of the test cases:

In the first test case, if the computer chooses the subsequence [1 2 3], our hero can replace the numbers 0 and 3. In this case, he can actually win for every chosen contiguous subsequence, except the entire array.

In the second test case, if the computer chooses the entire array [3 7 0 4 6 1 5 2] as the lit up subsequence, our hero can’t change the XOR of the subsequence (which is 0), no matter which two elements are swapped.

https://www.luogu.org/problemnew/show/P4443

题目大意：：

给定一个正整数M,和0到2^M-1共2^M个数的排列，计算机先选择一段连续的子串，你必须选两个不同的元素出来，交换它们的位置，使得计算机选出来的子串在交换后的元素的异或值为2^M-1,如果可以，你将获胜。现在请你统计有多少种不同的选法，你可以获胜。

就是叫你求出所有满足条件的区间的个数。。。

官方标程是线段树 + ！@#￥%…… 我看不懂......(orz洛谷大佬写的线段树)......

线段树：： //这个真的看不懂QAQ......

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

#define ms(s, n) memset(s, n, sizeof(s))
#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define FORd(i, a, b) for (int i = (a) - 1; i >= (b); i--)
#define FORall(it, a) for (__typeof((a).begin()) it = (a).begin(); it != (a).end(); it++)
#define sz(a) int((a).size())
#define present(t, x) (t.find(x) != t.end())
#define all(a) (a).begin(), (a).end()
#define uni(a) (a).erase(unique(all(a)), (a).end())
#define pb push_back
#define pf push_front
#define mp make_pair
#define fi first
#define se second
#define prec(n) fixed<<setprecision(n)
#define bit(n, i) (((n) >> (i)) & 1)
#define bitcount(n) __builtin_popcountll(n)
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int, int> pi;
typedef vector<int> vi;
typedef vector<pi> vii;
;
;
const int INF = (int) 1e9;
const ll LINF = (ll) 1e18;
);
;
inline ll gcd(ll a, ll b) {ll r; while (b) {r = a % b; a = b; b = r;} return a;}
inline ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
inline ll fpow(ll n, ll k, ; ) {) r = r * n % p; n = n * n % p;} return r;}
template< : ;}
template< : ;}
inline ll isqrt(ll k) {ll r = sqrt(k) + ; while (r * r > k) r--; return r;}
inline ll icbrt(ll k) {ll r = cbrt(k) + ; while (r * r * r > k) r--; return r;}
inline void addmod(int& a, int val, int p = MOD) {if ((a = (a + val)) >= p) a -= p;}
inline ) a += p;}
inline int mult(int a, int b, int p = MOD) {return (ll) a * b % p;}
inline , p);}
inline  : x > +EPS;}
inline int sign(ld x, ld y) {return sign(x - y);}
#define db(x) cerr << #x << " = " << (x) << " ";
#define endln cerr << "\n";

 << ;
int n;
int a[maxn];
int f[maxn];
];
int mn[maxn];
int mx[maxn];

][maxn << ];
void updmn(int st[], int p, int val) {
    ; ) p >>= , st[p] = min(st[p << ], st[p <<  | ]);
}
int querymn(int st[], int l, int r) {
    int res = INF;
    ; l < r; l >>= , r >>= ) {
        ) chkmin(res, st[l++]);
        ) chkmin(res, st[--r]);
    }
    return res;
}
void updmx(int st[], int p, int val) {
    ; ) p >>= , st[p] = max(st[p << ], st[p <<  | ]);
}
int querymx(int st[], int l, int r) {
    int res = -INF;
    ; l < r; l >>= , r >>= ) {
        ) chkmax(res, st[l++]);
        ) chkmax(res, st[--r]);
    }
    return res;
}

void solve() {
    cin >> n; n =  << n;
    FOR(i, , n) mn[i] = INF, mx[i] = -INF;
    FOR(i, , n) {
        cin >> a[i];
        chkmin(a[i], a[i] ^ n - );
        chkmin(mn[a[i]], i);
        chkmax(mx[a[i]], i);
    }
    ) {
        cout <<  << "\n";
        return;
    }
    FOR(i, , n) {
        updmn(st[], i, mn[a[i]]);
        updmx(st[], i, mx[a[i]]);
    }
    ;
    FOR(i, , n) {
        int k = a[i];
        if (mx[k] == i) {
            f[i] = -;
            ], mn[k], mx[k]);
            if (j >= mn[k]) {
                f[i] = j;
            }
            else {
                f[i] = f[mx[a[j]]];
            }
             && querymx(st[], f[i], i) <= i) {
                ;
                 & ;
                dp[i][par] = ;
                if (f[i]) {
                    dp[i][par] += dp[f[i] - ][];
                    dp[i][par ^ ] += dp[f[i] - ][];
                }
                res -= dp[i][];
            }
        }
    }
    res += () / ;
    cout << res << "\n";
}

int main(int argc, char* argv[]) {
    ios_base::sync_with_stdio(), cin.tie();
    ) {
        assert(freopen(argv[], "r", stdin));
    }
    ) {
        assert(freopen(argv[], "wb", stdout));
    }
    solve();
    ;
}//标程....

但是我只会写写被数据卡到飞起的hash......怎么办？一个hash不行就2个嘛......

hash : : //代码极丑.......毕竟是蒟蒻.......体谅下

//hash 有毒, 双hash才可过 ;
#include <cstdio>
#include <map>
using namespace std;

typedef long long ll;
 << ;
map<pair<ll, ll>, ];
; return c = 1ll * c * 15315ll % 0x7fffffff; }
], pos[N + ], f1[N + ], f2[N + ], r1[N + ], r2[N + ];

int main() {
    scanf( << n; ll ans = 1ll * c * (c + ) / 2ll;
    ; i <= c; i ++) scanf("%d", a + i), pos[a[i]] = i;

    ; i <= c / ; i ++) r1[pos[i]] = r1[pos[i ^ (c-)]] =  rand(), r2[pos[i]] = r2[pos[i ^ (c-)]] = rand();

    ; i <= c; i ++) f1[i] = f1[i-] ^ r1[i], f2[i] = f2[i-] ^ r2[i];

    ; i <= c; i ++) ans -= p[i % ][make_pair(f1[i], f2[i])], p[i % ][make_pair(f1[i], f2[i])] ++;

    printf(n ==  ? " : "%lld\n", ans);
}

证明如下：：

　　设 k = (1 << M) - 1; pos[] 表示值为 x 的数的位置；

　　首先考虑区间长度为奇数的时候，设区间的xor值为 A，则将其中某个数换走后的xor值为 A ^a[i]，那么要使区间 xor值为 k，则需要换一个 a[j] = k ^ A ^a[i] (这不是废话么......)，设 g = k ^ A(这是个定值)

假设该区间不合法，则有 pos[g ^ a[1]] ∈ [l, r], pos[g ^ a[2]] ∈ [l, r]......，因为 a[i] != a[j]，所以g ^ a[i] != g ^a[j](又是废话),

所以g ^ a[1] ^ g ^ a[2] ^ .... g ^ a[n] = a[1] ^ a[2] ^ ...^a[n]，所以g ^ a[1] ^ a[2] ^ ... ^a[n] == a[1] ^ a[2] ^ ... ^ a[n]，所以当前情况 g = 0。。。又 g = k ^ A，所以 k == A (与假设矛盾)。所以当区间长度为奇数时一定是

合法区间。

　　好，现在开始考虑区间长度为偶数时先讨论区间长度 / 2 为奇数时

　　　　同样，如果为不合法区间 (假设) ,则有 a[ i ] ^ g = a[ j ], pos[ a[ i ] ] ∈ [l, r], pos[ a[ j ] ] ∈ [l, r]; 又因为 g = a[ i ] ^ a[ j ]; 所以最后区间的 xor 值一定为 g (自己想一想)。

　　　　所以 A = a[i] ^ a[j]。 g = K ^ A = a[i] ^a[j]; k ^ a[i] ^ a[j] = a[i] ^a[j]。(好像在哪里见过。。。。）

　　现在讨论区间长度为偶数且区间长度 / 2 为偶数的情况

　　　　同上，区间的 xor 值 A 为 0(不想解释了，反正上面看懂了也很简单), 则有 g = K(g = k ^ A), 所以 a[ i ] ^ a[ j ] = k (对于任意a[i]都只有一个a[j]与之异或等于K,不想证明了), 所以到这里我们就把所有不合法区间的条件求出了。

　　总结下，现在我们知道了区间的总数为 (k + 2) * (k + 1) / 2,区间长度为4的倍数且其中的数都是配♂对的数的区间不合法，减去，双hash乱搞ok(一个hash让我Wa到怀疑人生）；

证明完毕。。。（我语文很差，可能有的地方表达有误。谁看懂标程的可以说下否？）