2017 Multi-University Training Contest

题意：给你n个数，其中每个区间的权值就是这个区间内不同数的个数除以区间长度，求最小的权值

令size(l, r)表示区间[l, r]内不同数的个数，那么就是要求min(size(l, r)/(r-l+1))  (1<=l<=n)(1<=r<=n)

可以二分答案ans = size(l, r)/(r-l+1)

将式子转化一下：size(l, r)+ans*l=ans*(r+1)

其中size(l, r)+ans*(l-1)很明显可以用线段树维护最小值

然后二分每次判定min(size(l, r)+ans*(l-1))是否小于ans*t就好了

又是道经典题，线段树的处理方式和下面两题几乎一模一样

http://blog.csdn.net/jaihk662/article/details/76559758

http://blog.csdn.net/jaihk662/article/details/76451307

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int n, pre[60005], a[60005], pos[60005];double tre[245424], temp[245424];void Lazy(int l, int r, int x){int m;m = (l+r)/2;tre[x*2] += temp[x];tre[x*2+1] += temp[x];if(l!=m)temp[x*2] += temp[x];if(r!=m+1)temp[x*2+1] += temp[x];temp[x] = 0;}void Update(int l, int r, int x, int a, int b, double c){int m;if(l>=a && r<=b){tre[x] += c;if(l!=r)temp[x] += c;return;}if(temp[x])Lazy(l, r, x);m = (l+r)/2;if(a<=m)Update(l, m, x*2, a, b, c);if(b>=m+1)Update(m+1, r, x*2+1, a, b, c);tre[x] = min(tre[x*2], tre[x*2+1]);}double Query(int l, int r, int x, int a, int b){int m;double ans = 2*n+1;if(l>=a && r<=b)return tre[x];if(temp[x])Lazy(l, r, x);m = (l+r)/2;if(a<=m)ans = min(ans, Query(l, m, x*2, a, b));if(b>=m+1)ans = min(ans, Query(m+1, r, x*2+1, a, b));return ans;}int Jud(double val){int i;memset(tre, 0, sizeof(tre));memset(temp, 0, sizeof(temp));for(i=1;i<=n;i++){Update(1, n, 1, pre[i]+1, i, 1);Update(1, n, 1, i, i, val*i);if(Query(1, n, 1, 1, i)<=val*(i+1))return 1;}return 0;}int main(void){int T, i;double l, r, m;scanf("%d", &T);while(T--){scanf("%d", &n);memset(pos, 0, sizeof(pos));for(i=1;i<=n;i++){scanf("%d", &a[i]);pre[i] = pos[a[i]];pos[a[i]] = i;}l = 0, r = 1;while(r-l>0.00005){m = (l+r)/2;if(Jud(m))r = m;elsel = m;}printf("%f\n", r);}return 0;}