KYEREMAL-BZOJ2038-[2009 Country Training Team-small Z socks (hose)-MO Team algorithm

bzoj2038-[2009 Country Training team-small Z socks (hose)

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Notice: Save the season Happy & another year of World final data, thank you &OJ Exam Test Breakthrough 3000 Mark! 2038: [2009 countries Training team] small Z socks (hose) Time limit: Sec Memory Limit: 259 MB
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As a rambling person, little Z spends a lot of time every morning finding a pair to wear from a bunch of colorful socks.

Eventually, one day, littleZCan no longer endure this annoying to find socks process, so he decided to resign...
In detail, the smallZPut thisNJust socks from1ToNNumber. Then from the numberLToR (LAlthough smallZDo not care about two socks is not a complete pair, and even do not care whether two socks is a left and right, he is very concerned about the color of socks, after all, wearing two different colors of socks will be very embarrassing.
Your job is to tell the smallZ。 How much he had the chance to draw two socks that were just 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 consists of 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, in which CI denotes the color of the sock, and the same color is represented by the same number. Next m line, two positive integer l per line. R indicates a query.

Output

Include M-line. For each query in a row, the output fraction of A/b represents the same probability of randomly extracting two of the same socks color from the range [l,r] of the query.

If the probability is 0, the output is 0/1. Otherwise the output A/b must be a minimal fraction. (see example)

Sample Input6 4
1 2 3 3 3 2
2 6
1 3
3 5
1 6
Sample Output2/5
0/1
1/1
4/15
"Example Explanation"
Question 1: Total C (5,2) = 10 possible, two out of 2 of which there are 1 possible, the extraction of two 3 is 3 possible, the probability is (1+3)/10=4/10=2/5.

Question 2: Total C (3,2) = 3 possible. Can not draw the same color socks, the probability of 0/3=0/1.
Inquiry 3: Total C (3,2) = 3 possible, are extracted two 3, the probability is 3/3=1/1.
Note: The above C (A, b) represents the number of combinations, the combination of C (A, B) is equivalent to the selection of B in a different item number of selection scheme.
"Data size and conventions"
30% of the data are n,m≤5000;
60% of the data are n,m≤25000;
100% of the data are n,m≤50000. 1≤l < R≤n,ci≤n.

HINT

Source

All rights reserved by: Mo Tao

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The main idea: give a bunch of numbers, each time asked an interval, the interval to take two number of the same probability.

Analysis: Orz Mo Tao, MO team to the various range of inquiries

Give a few blogs about the MO team algorithm:

Very popular and intuitive to understand the algorithm

The specific

Rigorous plane dot Manhattan minimum distance spanning tree thesis

A more rigorous English paper

Code

#include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <cstring > #include <algorithm>using namespace std; #define REP (I, L, R) for (LL i = (l); (i) <= (R); ((i) + +)) #define REP (I, L, R) for (LL i = (l); (i) >= (R); ((i)-))  #define LL Long Long#define maxn 1000010LL N, M, COL[MAXN], LEN[MAXN], K, C[MAXN], POS[MAXN], FZ, FM, Minc, maxc;struct QR {LL L, r, Num;} A[maxn];struct Answer {ll x, y, num;} ans[maxn];inline ll min (ll a, ll b) {return a<b? A:b;} inline ll Max (ll A, ll b) {return a>b?
 A:B;} inline BOOL CMP (QR A, QR b) {return POS[A.L]&LT;POS[B.L] | | (POS[A.L]==POS[B.L] && A.R&LT;B.R);} Inline ll com (ll N) {return (n-1) *N/2;} Inline answer MP (ll X, ll y, ll num) {answer t; t.x = x, T.y = y, t.num = num; return t;} inline bool Cmp2 (answer A, answer b) {return a.num < B.num;} inline ll gcd (ll A, ll b) {return!b? A:GCD (b, a%b);} inline void Update (LL &l, LL &r, LL L, LL R, LL &fz, ll &AMP;FM) {//[L, R]-[L, R] if (L < L) REP (    I, L-1, L) FZ + = C[col[i]], fm + = r-l+1, c[col[i]]++, l--;    else if (L > L) Rep (I, L, l-1) FZ + =-c[col[i]] + 1, FM + L-R, C[col[i]]--, l++;    if (R > R) Rep (I, r+1, r) FZ + = C[col[i]], fm + = r-l+1, c[col[i]]++, r++; else if (R < R) REP (I, R, r+1) FZ + =-c[col[i]] + 1, fm + = L-r, C[col[i]]--, r--;}    inline void block () {K = sqrt (n);    Rep (i, 1, K-1) len[i] = K;    Len[k] = n-(K-1) *k;    Rep (i, 1, K) Rep (J, (I-1) *k + 1, (i-1) *k + len[i]) pos[j] = i; Sort (a+1, a+1+m, CMP);} inline void xx () {   memset (c, 0, sizeof (c));    LL L = a[1].l, R = A[1].R;    Rep (i, L, R) c[col[i]]++;    Rep (i, Minc, maxc) FZ + = (c[i] * (c[i]-1)/2);    fm = COM (r-l+1);    ANS[1] = MP (FZ, FM, A[1].num); Rep (I, 2, m) update (L, R, A[I].L, A[I].R, FZ, FM), Ans[i] = MP (FZ, FM, a[i].num);}    int main () {minc = maxn, maxc = 0;    CIN >> n >> m;    Rep (i, 1, N) scanf ("%lld", col + i), MAXC = Max (MAXC, Col[i]), minc = min (Minc, col[i]);    Rep (i, 1, m) scanf ("%lld%lld", &AMP;A[I].L, &AMP;A[I].R), a[i].num = i;    Block ();    XX ();    Sort (ans+1, ans+1+m, CMP2);        Rep (i, 1, m) n = gcd (ans[i].x, Ans[i].y), printf ("%lld/%lld\n", ans[i].x/n, ans[i].y/n); return 0;}

KYEREMAL-BZOJ2038-[2009 Country Training Team-small Z socks (hose)-MO Team algorithm