[bzoj2038 [2009国家集训队]小Z的袜子(hose)] 莫队算法

题意：Q个询问，每个询问给定区间[L,R]，求从里面任选两个数相同的概率。

思路：莫队算法。用一个cnt数组记录当前区间每种数的个数，区间变化为1时O(1)的维护cnt数组，并可以O(1)的得到当前区间中与当前数相同的数的个数。

 

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#pragma comment(linker, "/STACK:10240000")
#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

#define X                   first
#define Y                   second
#define pb                  push_back
#define mp                  make_pair
#define all(a)              (a).begin(), (a).end()
#define fillchar(a, x)      memset(a, x, sizeof(a))
#define copy(a, b)          memcpy(a, b, sizeof(a))

typedef long long ll;
typedef pair<int, int> pii;
typedef unsigned long long ull;

//#ifndef ONLINE_JUDGE
void RI(vector<int>&a,int n){a.resize(n);for(int i=0;i<n;i++)scanf("%d",&a[i]);}
void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R>
void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?1:-1;
while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T>
void print(const T t){cout<<t<<endl;}template<typename F,typename...R>
void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T>
void print(T*p, T*q){int d=p<q?1:-1;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;}
//#endif
template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);}
template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);}

const double PI = acos(-1.0);
const int INF = 1e9 + 7;
const double EPS = 1e-8;

/* -------------------------------------------------------------------------------- */

const int maxn = 5e4 + 7;

pair<pii, int> qry[maxn];
int a[maxn], block, cnt[maxn];
pair<ll, ll> ans[maxn];

ll gcd(ll a, ll b) {
    return b? gcd(b, a % b) : a;
}
bool cmp(const pair<pii, int> &a, const pair<pii, int> &b) {
    int lb = a.X.X / block, rb = b.X.X / block;
    return lb == rb? a.X.Y < b.X.Y : lb < rb;
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
#endif // ONLINE_JUDGE
    int n, m;
    while (cin >> n >> m) {
        block = (int)sqrt(n * 1.0 + 0.5);
        for (int i = 0; i < n; i ++) {
            scanf("%d", a + i);
        }
        for (int i = 0; i < m; i ++) {
            scanf("%d%d", &qry[i].X.X, &qry[i].X.Y);
            qry[i].X.X --;
            qry[i].X.Y --;
            qry[i].Y = i;
        }
        sort(qry, qry + m, cmp);
        fillchar(cnt, 0);
        int lastl = 0, lastr = 0;
        ll lastans = 0;
        cnt[a[0]] ++;
        for (int i = 0; i < m; i ++) {
            while (qry[i].X.X < lastl) lastans += cnt[a[-- lastl]] ++;
            while (qry[i].X.Y > lastr) lastans += cnt[a[++ lastr]] ++;
            while (qry[i].X.X > lastl) lastans -= -- cnt[a[lastl ++]];
            while (qry[i].X.Y < lastr) lastans -= -- cnt[a[lastr --]];
            ll n = qry[i].X.Y - qry[i].X.X + 1, buf = n * (n - 1) / 2;
            ll g = gcd(lastans, buf);
            if (lastans) ans[qry[i].Y] = mp(lastans / g, buf / g);
            else ans[qry[i].Y] = mp(0, 1);
        }
        for (int i = 0; i < m; i ++) {
            printf("%lld/%lld\n", ans[i].X, ans[i].Y);
        }
    }
    return 0;
}
posted @ 2015-08-15 21:50  jklongint  阅读(180)  评论(0)  编辑  收藏  举报