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 ", ans[i].X, ans[i].Y); } } return 0; } |