HDU

HDU - 5857 

题意： 给一个已经排好序的序列，然后给你两个区间，问这两个区间的中位数是多少。

思路：直接模拟一定超时，所以得考虑不同的情况，：）贼容易错！中间变量会超long long

不同的情况就是区间不相交和区间相交，区间相交的那一段每个元素有两个。

代码：

#include <cstdio>#include <iostream>#include <algorithm>#include <cstring>#include <queue>#include <set>using namespace std;#define ll long longconst int maxn = 1e5 + 10;ll D[maxn];int n,m;int Find(int a,int b,int c,int d,int id){    int l = c, r = min(b,d), last = max(b,d);    int cnum = (r-l+1)*2;    int pnum = l - a,  lnum = last - r, vnum = pnum + cnum + lnum;    if(id <= pnum)        return a + id - 1;    else if(id <= pnum + cnum)        return c + (id - pnum - 1)/2;    else return r + (id - pnum - cnum) ;}int Finda(int a,int b,int c,int d,int id){    int fnum = b - a +1;    if(id > fnum)        return c + id - fnum - 1;    else return a + id - 1;}void solve(int a,int b,int c,int d){    if(a > c) swap(a,c),swap(b,d);    int fnum = b-a+1, snum = d-c+1;    if(b < c)    {        int vnum = fnum + snum;        if(vnum % 2)        {            int id = vnum / 2 + 1;            int thi = Finda(a,b,c,d,id);            printf("%.1lf\n",(double)D[thi]);        }        else        {            int id1 = vnum / 2, id2 = id1 + 1;            int thi = Finda(a,b,c,d,id1), thj = Finda(a,b,c,d,id2);            printf("%.1f\n",(double)(D[thi] + D[thj])/2.0);        }    }    else    {        int vnum = c-a + (b-c+1)*2 + d-b;        if(vnum % 2)        {            int id = Find(a,b,c,d,vnum/2 + 1);            printf("%.1f\n",(double)D[id]);        }        else        {            int id1 = Find(a,b,c,d,vnum/2);            int id2 = Find(a,b,c,d,vnum/2 + 1);            printf("%.1f\n",(double)(D[id1]+D[id2])/2.0);        }    }}int main(){    int t,a,b,c,d;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);        for(int i = 1; i <= n; i++) cin >> D[i];        for(int i = 0; i < m; i++)        {            scanf("%d%d",&a,&b);            scanf("%d%d",&c,&d);            solve(a,b,c,d);        }    }return 0;}