Median

题目链接

题意：给你一个包含n个数的序列和m个询问，每个询问包含两个区间，求出这两个区间包含的所有数的中位数。

思路：分两种情况讨论即可，一种是两区间没有交集，一种是有交集。

代码：

/// He  renders  landscapes  with  great  skill  and  artistry.#define _CRT_SECURE_NO_WARNINGS#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <cstdlib>#include <cmath>#include <iterator>#include <cctype>#include <sstream>#include <string>#include <vector>#include <set>#include <map>#include <stack>#include <deque>#include <queue>#include <list>#include <functional>#include <ctime>#include <bitset>//#pragma comment(linker, "/STACK:102400000, 102400000")#define debug     puts("+******************************************************+")#define Min(a, b) ( (a < b) ? a : b )#define Max(a, b) ( (a > b) ? a : b )#define lc         o<<1#define rc         o<<1|1#define lson       L, M, lc#define rson       M + 1, R, rc#define mem0(x)   memset(x, 0, sizeof x)#define mem1(x)   memset(x, -1, sizeof x)#define memf(x)   memset(x, false, sizeof x)#define pb        push_back#define pf        push_front#define LB        lower_bound#define UB        upper_bound#define PQ        priority_queue#define fr(x)     freopen("x", "r", stdin )#define fw(x)     freopen("x", "w" , stdout)#define all(a)    a.begin(), a,end()#define X         first#define Y         second#define MP        make_pair#define Abs(x)    ( x >= 0 ) ? x : ( -x )#define MAXS      50000 + 8#define MAXT      10000 + 8#define MAXL      500000 + 8#define INF       0x3f3f3f3f#define INFL      1000000000000000000#define inf       -(1<<30)#define EPS       1e-10#define PI        acos(-1.0)#define sqr(x)    (x * x)using namespace std;typedef long long          LL;typedef unsigned long long uLL;typedef double             DB;typedef long double        LD;typedef pair<int, int >   pii;typedef pair<LL, LL>       pll;const int MOD   = 1e9;const int N     = 1e5 + 8;const int maxn  = 1e3 + 8;const int dx[]  = { -1, 1,  0, 0 };const int dy[]  = {  0, 0, -1, 1 };struct Node{    LL x;    int id;    char ch;    bool operator < ( const Node & n) const    {        return x < n.x;    }} no[N];int sum;int l1, r1, l2, r2;int n, m;double a[N];double solve(int len){     if ( r1 < l2 ) {          if ( l1  + len - 1  <= r1 ) return a[l1  + len - 1];          else {              int t = l1 + len - 1 - r1;              return a[ l2  + t - 1 ];          }     } else {          if ( l1  + len - 1  < l2  ) return a[l1  + len - 1];          else if (  l2 - 1 +  (r1 - l2 + 1 ) * 2 <= l1 + len - 1 ) {                  int t = l1  + len - 1 - r1;                  return a[l2 + t - 1];          } else{                 int t = l1 + len - 1 - (l2 - 1);                 if ( t & 1 ) return a[ l2 + t / 2 ];                 else return a[l2 + t / 2 - 1 ];          }     }}int len;int main(){    //freopen("codecoder.in", "r", stdin);    //freopen("out.txt", "w", stdout);    // ios::sync_with_stdio(false);    int t;    scanf("%d", &t);    while (t--)    {        scanf("%d%d", &n, &m);        for (int i = 1; i <= n; i++) scanf("%lf", &a[i]);        for (int i = 0; i < m; i++)        {            scanf("%d%d%d%d", &l1, &r1, &l2, &r2);            if ( l1 > l2 ) swap(l1, l2);            if ( r1 > r2 ) swap(r1, r2);            len = r1 - l1 + 1 + r2 - l2 + 1;            if ( len & 1 ) printf("%.1lf\n",  1.0 * solve(len / 2 + 1));            else printf("%.1lf\n",  0.5 * solve(len / 2) +  0.5 * solve(len / 2 + 1));        }    }    return 0;}

队内赛自己分情况讨论的时候把这题搞复杂了，改了好长时间没改出来，坑队友了。

附上比赛时的坑码（已AC）， 留个教训。

/// He  renders  landscapes  with  great  skill  and  artistry.#define _CRT_SECURE_NO_WARNINGS#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <cstdlib>#include <cmath>#include <iterator>#include <cctype>#include <sstream>#include <string>#include <vector>#include <set>#include <map>#include <stack>#include <deque>#include <queue>#include <list>#include <functional>#include <ctime>#include <bitset>//#pragma comment(linker, "/STACK:102400000, 102400000")#define debug     puts("+******************************************************+")#define Min(a, b) ( (a < b) ? a : b )#define Max(a, b) ( (a > b) ? a : b )#define lc         o<<1#define rc         o<<1|1#define lson       L, M, lc#define rson       M + 1, R, rc#define mem0(x)   memset(x, 0, sizeof x)#define mem1(x)   memset(x, -1, sizeof x)#define memf(x)   memset(x, false, sizeof x)#define pb        push_back#define pf        push_front#define LB        lower_bound#define UB        upper_bound#define PQ        priority_queue#define fr(x)     freopen("x", "r", stdin )#define fw(x)     freopen("x", "w" , stdout)#define all(a)    a.begin(), a,end()#define X         first#define Y         second#define MP        make_pair#define Abs(x)    ( x >= 0 ) ? x : ( -x )#define MAXS      50000 + 8#define MAXT      10000 + 8#define MAXL      500000 + 8#define INF       0x3f3f3f3f#define INFL      1000000000000000000#define inf       -(1<<30)#define EPS       1e-10#define PI        acos(-1.0)#define sqr(x)    (x * x)using namespace std;typedef long long          LL;typedef unsigned long long uLL;typedef double             DB;typedef long double        LD;typedef pair<int, int >   pii;typedef pair<LL, LL>       pll;const int MOD   = 1e9;const int N     = 1e5 + 8;const int maxn  = 1e3 + 8;const int dx[]  = { -1, 1,  0, 0 };const int dy[]  = {  0, 0, -1, 1 };struct Node{    LL x;    int id;    char ch;    bool operator < ( const Node & n) const    {        return x < n.x;    }} no[N];int sum;int l1, r1, l2, r2;int n, m;double a[N];int z;void solve1(){   // debug;    z = sum / 2;    int zh = l1 + z; //printf("%d------\n", zh);    if ( r1 < l2 )    {        if ( zh <= r1 ) printf("%.1lf\n", a[zh] );        else        {            int t = zh - r1;            printf("%.1lf\n", a[ l2 + t - 1  ] );        }    }    else if ( r1 >= l2 && r1 <= r2 )    {        int t = r1 - l2 + 1;        int tz = l2 - l1;        int tx = z  + 1 - tz; ///个数        if ( zh < l2 ) { }        else {           if ( l1 + tz - 1 + (r1 - l2 + 1) * 2  >= zh )             zh = l2 + ( (tx % 2 == 0) ? (tx / 2 - 1) : (tx / 2) );           else {             int tt = zh - r1;             zh = l2 + tt - 1;           }        }        //printf("%d %d %d++++\n", t, tx, zh);        printf("%.1lf\n", a[zh] );    }    else if ( r1 > r2 ){        int t = r1 - l2 + 1;        int tz = l2 - l1;        int tx = zh - tz;        if ( l1 + tz  - 1 + (r2 - l1 + 1) * 2 < zh ) {             int tt = zh  - l1 + 1 - tz - (r2 - l2 + 1) * 2;             zh = r2 + tt - 1;        } else {             zh = l2 + ( (tx % 2 == 0) ? (tx / 2 - 1) : (tx / 2) );        }         printf("%.1lf\n", a[zh] );    }}double f(int zh){   // printf("%d+++++++\n", zh);     int t = r1 - l2 + 1;        int tz = l2 - l1;        int tx = zh - l1  + 1 - tz; ///个数        if ( zh < l2 ) {   }        else {           if ( l1 + tz - 1 + (r1 - l2 + 1) * 2  >= zh )             zh = l2 + ( (tx % 2 == 0) ? (tx / 2 - 1) : (tx / 2) );           else {             int tt = zh - r1;             zh = l2 + tt - 1;           }        }        return a[zh];        //printf("%d %d %d++++\n", t, tx, zh);       // printf("%.1lf\n", a[zh] );}double ff(int zh){   // printf("%d+++++++\n", zh);    int t = r1 - l2 + 1;        int tz = l2 - l1;        int tx = zh - tz;        if ( zh < l2) { }        else {          if ( l1 + tz  - 1 + (r2 - l1 + 1) * 2 < zh ) {             int tt = zh  - l1 + 1 - tz - (r2 - l2 + 1) * 2;             zh = r2 + tt - 1;          } else {             zh = l2 + ( (tx % 2 == 0) ? (tx / 2 - 1) : (tx / 2) );          }        }        return a[zh];}void solve2(){    z = sum / 2;    int zh = l1 + z - 1;    if ( r1 < l2 )    {        if ( zh < r1 ) {                printf("%.1lf\n", ( a[zh] + a[zh +  1] )/ 2 );        }        else if ( zh == r1) {               printf("%.1lf\n", ( a[zh] + a[l2] ) / 2 );        }        else {            int t = zh - r1;            zh = l2 + t - 1;            printf("%.1lf\n", ( a[zh] + a[zh +  1] )/ 2 );        }    }    else if ( r1 >= l2 && r1 <= r2)    {        double pp = f(zh);        double qq = f(zh + 1);        //printf("%lf %lf+++++++++++\n", pp, qq);        printf("%.1lf\n", ( pp + qq )  / 2 );    }    else if ( r1 > r2 ){        //debug;        double pp = ff(zh);        double qq = ff(zh + 1);       // printf("%lf %lf+++++++++++\n", pp, qq);         printf("%.1lf\n", ( pp + qq )  / 2 );    }}int main(){    //freopen("codecoder.in", "r", stdin);    //freopen("out.txt", "w", stdout);    // ios::sync_with_stdio(false);    int t;    scanf("%d", &t);    while (t--)    {        scanf("%d%d", &n, &m);        for (int i = 1; i <= n; i++) scanf("%lf", &a[i]);        for (int i = 0; i < m; i++)        {            scanf("%d%d%d%d", &l1, &r1, &l2, &r2);            if (l1 > l2) swap(l1, l2);            if (r1 > r2) swap(r1, r2);            sum = r1 - l1 + 1 + r2 - l2 + 1;            if ( sum & 1 ) solve1();            else solve2();        }    }    return 0;}