HYSBZ 2038 small Z Socks (hose) (Mo team algorithm)

Test instructions: Chinese question.

Analysis: The very famous MO team algorithm, the first expression of this probability, should be the molecule: C (x1, 2) + C (x2, 2) + C (X3, 2) + ... + c (xn, 2) Denominator: C (n, 2), and then the expression of fractions, [X1 (x1-1) +x2 (x2-1) +...+XN (xn-1)]/(n (n-1)) then the simplification is obtained (Sigma (XI*XI)-N)/(n (n-1)), then the operation of each interval, offline, so that the sequence is divided into sqrt (n) block, and then with two pointers, the calculation of the data.

Be careful not to use i64d, the words wa to die ...

The code is as follows:

#pragma COMMENT (linker, "/stack:1024000000,1024000000") #include <cstdio> #include <string> #include < cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include < queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include < cmath> #include <stack> #include <assert.h> #include <sstream> #define DEBUG () puts ("++++"); Define GCD (A, B) __gcd (A, b) #define Lson l,m,rt<<1#define Rson m+1,r,rt<<1|1#define freopenr freopen ("In.txt "," R ", stdin) #define FREOPENW freopen (" OUT.txt "," w ", stdout) using namespace Std;typedef long long ll;typedef unsigned lo ng Long Ull;typedef pair<int, int> p;const int inf = 0x3f3f3f3f;const LL LNF = 0xffffffffffll;const double INF = 0x3  F3f3f3f3f3f;const Double PI = ACOs ( -1.0); const double EPS = 1e-6;const int maxn = 50000 + 10;const int mod = 1e9 + 7;const int dr[] = {-1, 0, 1, 0};const int dc[] = {0, 1, 0, -1};const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1 101 "," 1110 "," 1111 "};int N, m;const int mon[] = {0, 0,, +, +,,, 31};const int monn[] , d, N, 0, D, D, D, D,, D,, 31};inline bool is_in (int r, int c) {return R >= && r < N & amp;& C >= 0 && C < m;}  LL ANSX[MAXN], Ansy[maxn];int POS[MAXN], val[maxn];int cnt[maxn];struct node{int L, r, id;  BOOL operator < (const Node &p) const{return pos[l] < POS[P.L] | | pos[l] = = POS[P.L] && r < P.R; }}; Node A[MAXN];  ll x, y;void update (int l, int ok) {x-= Cnt[val[l]] * (LL) cnt[val[l]];  CNT[VAL[L]] + = OK;  x + = Cnt[val[l]] * (LL) cnt[val[l] "-OK; Y + = OK;}  void update (int i) {ansx[i] = x; Ansy[i] = y * (y-1);}  int main () {scanf ("%d%d", &n, &m);  for (int i = 1; I <= n; ++i) scanf ("%d", val+i); for (int i = 0; i < m; ++i) {scanf ("%d%d", &a[i].L, &AMP;A[I].R);  A[i].id = i;  } int t = sqrt (n + 0.5);  for (int i = 1; I <= n; ++i) pos[i] = i/t;  Sort (A, A + M);    for (int l = 1, i = 0, r = 0; i < m; ++i) {int L = A[I].L, r = A[I].R;    while (L < L) Update (l++,-1);    while (L > L) Update (--l, 1);    while (R < R) Update (++R, 1);    while (R > R) Update (r--,-1);  Update (a[i].id);    } for (int i = 0; i < m; ++i) {LL g = gcd (Ansx[i], ansy[i]);  printf ("%lld/%lld\n", ansx[i]/g, ansy[i]/g); } return 0;}

　　

HYSBZ 2038 small Z Socks (hose) (Mo team algorithm)