1 //Accepted    2665    703MS    16388K    2972 B    G++  2 #include 
      
         3 #include 
       
          4 using namespace std;  5 #define M 100001  6 #define LL(x)     ((x)<<1)  7 #define RR(x)     ((x)<<1|1)  8 struct Seg_Tree{  9     int left,right; 10     int mid() { 11         return (left + right) >> 1; 12     } 13 }tt[M*4]; 14 int len; 15 int sorted[M]; 16 int toLeft[20][M]; 17 int val[20][M]; 18   19 void build(int l,int r,int d,int idx) { 20     tt[idx].left = l; 21     tt[idx].right = r; 22     if(tt[idx].left == tt[idx].right)    return ; 23     int mid = tt[idx].mid(); 24     int lsame = mid - l + 1;//lsame表示和val_mid相等且分到左边的 25     for(int i = l ; i <= r ; i ++) { 26         if(val[d][i] < sorted[mid]) { 27             lsame --;//先假设左边的数(mid - l + 1)个都等于val_mid,然后把实际上小于val_mid的减去 28         } 29     } 30     int lpos = l; 31     int rpos = mid+1; 32     int same = 0; 33     for(int i = l ; i <= r ; i ++) { 34         if(i == l) { 35             toLeft[d][i] = 0;//toLeft[i]表示[ tt[idx].left , i ]区域里有多少个数分到左边 36         } else { 37             toLeft[d][i] = toLeft[d][i-1]; 38         } 39         if(val[d][i] < sorted[mid]) { 40             toLeft[d][i] ++; 41             val[d+1][lpos++] = val[d][i]; 42         } else if(val[d][i] > sorted[mid]) { 43             val[d+1][rpos++] = val[d][i]; 44         } else { 45             if(same < lsame) {
    //有lsame的数是分到左边的 46                 same ++; 47                 toLeft[d][i] ++; 48                 val[d+1][lpos++] = val[d][i]; 49             } else { 50                 val[d+1][rpos++] = val[d][i]; 51             } 52         } 53     } 54     build(l,mid,d+1,LL(idx)); 55     build(mid+1,r,d+1,RR(idx)); 56 } 57   58 int query(int l,int r,int k,int d,int idx) { 59     if(l == r) { 60         return val[d][l]; 61     } 62     int s;//s表示[ l , r ]有多少个分到左边 63     int ss;//ss表示 [tt[idx].left , l-1 ]有多少个分到左边 64     if(l == tt[idx].left) { 65         s = toLeft[d][r]; 66         ss = 0; 67     } else { 68         s = toLeft[d][r] - toLeft[d][l-1]; 69         ss = toLeft[d][l-1]; 70     } 71     if(s >= k) {
    //有多于k个分到左边,显然去左儿子区间找第k个 72         int newl = tt[idx].left + ss; 73         int newr = tt[idx].left + ss + s - 1;//计算出新的映射区间 74         return query(newl,newr,k,d+1,LL(idx)); 75     } else { 76         int mid = tt[idx].mid(); 77         int bb = l - tt[idx].left - ss;//bb表示 [tt[idx].left , l-1 ]有多少个分到右边 78         int b = r - l + 1 - s;//b表示 [l , r]有多少个分到右边 79         int newl = mid + bb + 1; 80         int newr = mid + bb + b; 81         return query(newl,newr,k-s,d+1,RR(idx)); 82     } 83 } 84   85 int main() { 86     int T; 87     scanf("%d",&T); 88     while(T --) { 89         int n , m; 90         scanf("%d%d",&n,&m); 91         for(int i = 1; i <= n; i++){ 92             scanf("%d",&val[0][i]); 93             sorted[i] = val[0][i]; 94         } 95         sort(sorted + 1 , sorted + n + 1); 96         build(1,n,0,1); 97         while(m --) { 98             int l,r,k; 99             scanf("%d%d%d",&l,&r,&k);100             printf("区间第K小：%d\n",query(l,r,k,0,1));101             k = r - l + 2 - k;102             printf("区间第K大：%d\n",query(l,r,k,0,1));103         }104     }105     return 0;106 }