# P1494 [national training team] Small Z so

Source: Internet
Author: User

Secretly cut off the team's template question...

This question is also a very common MO team template question.

Asking you for a probability is actually asking you\ (\ Sum {\ frac {CNT [I] \ times (CNT [I]-1)} {2 }}\). If the denominator is human, please. After finding two things, you can divide them.

The numerator is actually a maintenance method in the same question of "B's inquiry.

The answer varies with each number. Then we will lose our current contribution to the answer, update the CNT array, and add a new contribution to the answer, so we can maintain it.

Code:

``# Include <cstdio> # include <cmath> # include <algorithm> # define ll long longconst int maxn = 50005; struct query {ll l, R, ID ;} ques [maxn]; ll belong [maxn]; ll out1 [maxn], out2 [maxn]; ll CNT [maxn]; ll a [maxn]; ll n, m; ll l = 1, r = 0, ANS = 0; bool CMP (query a, query B) {If (belong [. l] = belong [B. l]) return. r <B. r; return belong [. l] <belong [B. l];} ll read () {ll ans = 0, S = 1; char CH = getchar (); While (CH> '9' | ch <'0') {If (CH = '-') S =-1; CH = getchar ();} while (CH> = '0' & Ch <= '9') ans = (ANS <3) + (ANS <1) + CH-'0 ', ch = getchar (); Return S * ans;} void add (ll x) {ANS-= (CNT [x] * (CNT [x]-1)/2; CNT [x] ++; ans + = (CNT [x] * (CNT [x]-1)/2;} void del (ll x) {ANS-= (CNT [x] * (CNT [x]-1)/2; CNT [x] --; ans + = (CNT [x] * (CNT [x]-1)/2;} void moqueue () {ll block = SQRT (n); (INT I = 1; I <= N; I ++) belong [I] = (I-1)/block + 1; STD: Sort (ques + 1, ques + m + 1, CMP); For (INT I = 1; I <= m; I ++) {If (ques [I]. L = ques [I]. r) {out1 [ques [I]. id] = 0, out2 [ques [I]. id] = 1; continue;} while (L <ques [I]. l) del (A [L ++]); While (L> ques [I]. l) add (A [-- l]); While (r> ques [I]. r) del (A [R --]); While (r <ques [I]. r) add (A [++ R]); out1 [ques [I]. id] = ans; out2 [ques [I]. id] = (R-l + 1) * (R-l)/2;} ll gcd (ll x, ll y) {return y = 0? X: gcd (Y, X % Y);} int main () {n = read (), M = read (); For (INT I = 1; I <= N; I ++) A [I] = read (); For (INT I = 1; I <= m; I ++) Ques [I]. L = read (), ques [I]. R = read (), ques [I]. id = I; moqueue (); For (INT I = 1; I <= m; I ++) {ll ans1 = out1 [I], ans2 = out2 [I]; ll G = gcd (ans1, ans2); ans1/= g; ans2/= g; printf ("% LLD/% LLD \ n", ans1, ans2 );} return 0 ;}``

p1494 [national training team] Xiao Z's so

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