[BZOJ2038] [2009 countries Training team] small Z socks (hose)

[BZOJ2038] [2009 countries Training team] small Z socks (hose)

Question Description

As a rambling man of life, littleZ every morning it takes a long time to find a pair to wear from a pile of colorful socks. Finally one day, Little Z can no longer endure this annoying to find socks process, so he decided to resign to fate ...
Specifically, little z numbered the n socks from 1 to N, and then from the number L to R (l though little Z doesn't care if two socks are a complete pair, even if two socks are left and right, he cares about the color of the socks, After all, wearing two different color socks will be very embarrassing.
Your task is to tell little Z how much he has the chance to draw two socks of the same color. Of course, little Z wants this probability to be as high as possible, so he may ask multiple (L,R) to facilitate his choice.

Input

The first line of the input file contains two positive integers n and M. n is the number of socks, M is the number of inquiries raised by small Z. The next line consists of n positive integer ci, where CI denotes the color of the sock, and the same color is represented by the same number. The next m line, two positive integer l for each line, R indicates a query.

Output

Contains m rows, for each query the output fraction of a line is a/b indicating the probability of randomly extracting two socks of the same color from the range [l,r] of the query. If the probability is 0 then output 0/1, otherwise the output of A/b must be the simplest fraction. (See examples)

Input example

6 4 1 2 3 3 3 2 2 6 1 3 3 5 1 6

Output example

2/50/11/14/

Data size and conventions

30% of the data are n,m≤5000;
60% of the data are n,m≤25000;
100% of the data in N,m≤50000,1≤l < R≤n,ci≤n.

Exercises

Notice the ci≤n, so you can record the number of each color, so that the answer is very easy to calculate. Further discovery: When we know the answer to the interval [L, R], it is easy to draw the answer to these intervals [L + 1, R], [L-1, R], [L, R-1], [L, R + 1], so that we can do it with Mo team.

#include <iostream> #include <cstdio> #include <algorithm> #include <cmath> #include <stack > #include <vector> #include <queue> #include <cstring> #include <string> #include <map > #include <set>using namespace std;const int buffersize = 1 << 16;char buffer[buffersize], *head, *TAIL;INL        ine Char Getchar () {if (Head = = Tail) {int L = fread (buffer, 1, buffersize, stdin);    Tail = (Head = buffer) + L; } return *head++;}    int read () {int x = 0, f = 1; char c = Getchar ();    while (!isdigit (c)) {if (c = = '-') f =-1; c = Getchar ();}    while (IsDigit (c)) {x = x * + C-' 0 '; c = Getchar ();} return x * f;} #define MAXN 50010#define LL long longint N, M, COL[MAXN], HAS[MAXN]; LL ANS[MAXN], bns[maxn];struct Que {int id, L, R;} QS[MAXN], Tq[maxn];bool Cmpl (Que A, Que b) {return A.L < B.L;} BOOL Cmpr (Que A, Que b) {return A.R < B.R;} ll GCD (ll A, ll b) {return!b? A:GCD (b, a% b);} int main () {n = Read (); m = read (); for (int i = 1; I <= n; i++) Col[i] = read (); for (int i = 1; I <= m; i++) qs[i].id = i, qs[i].l = Read (), q S[I].R = read (); sort (qs + 1, QS + M + 1, Cmpl);//for (int i = 1; I <= m; i++) printf ("%d%d%d\n", Qs[i].id, QS[I].L, QS [I].R); int mxl = QS[M].L, siz = (int) sqrt (MXL +. 5), NL = 1, nr = siz;for (int i = 1; i <= m;) {int J, K, x;for (j = i; J < m && nl <= qs[j].l && qs[j].l <= nr; j + +);//for (x = i; x <= J; x+ +) printf ("Here:%d%d%d\n", QS[X].L, QS[X].R, qs[x].id); sort (qs + I, QS + j + 1, cmpr); i = j + 1; NL = nr + 1; NR = nl + siz-1;} for (int i = 1; I <= m; i++) printf ("%d%d%d\n", Qs[i].id, QS[I].L, QS[I].R); nl = 1, nr = 0;  LL ans = 0;//has[col[1]] = 1;for (int i = 1; I <= m; i++) {int QL = QS[I].L, QR = Qs[i].r;while (qr > Nr) nr++, ans + = HAS[COL[NR], Has[col[nr]]++;while (QL < NL) nl--, ans + has[col[nl]], Has[col[nl]]++;while (QR < nr) has[col[nr]]-- , ans-= Has[col[nr]], nr--;while (QL > NL) has[col[NL]]--, ans-= has[col[nl]], nl++;int len = qr-ql + 1; Ans[qs[i].id] = ans; Bns[qs[i].id] = (LL) len * (LL) (len-1) >> 1ll;//printf ("%d%d%d%lld%d\n", Qs[i].id, QL, QR, ans, len);} for (int i = 1; I <= m; i++) {LL a = Ans[i], B = bns[i], g;if (!a) {puts ("0/1"); continue;} g = gcd (A, b); A/= g; b/= g;printf ("%lld/%lld\n", A, b);} return 0;}

[BZOJ2038] [2009 countries Training team] small Z socks (hose)