HDU 3473 劃分樹求某段區間的和

/*劃分樹：解題類型單一，給出10^5個點，給你 q 個查詢，每次查詢出[l,r] 中第 k 大的值。劃分樹：                                主系列  4 3 2 5 8 7 6 1 9                           排序後  1 2 3 4 5 6 7 8 9   中位元是 5                   劃左邊  4 3 2 5 1                   劃右邊  8 7 6 9             劃左邊  3 2 1  劃右邊 4 5            劃左邊  6 7 劃右邊 8 9              。。。。。。。                           。。。。。。 參考部落格：http://www.2cto.com/kf/201210/160552.html */#include<iostream>#include<algorithm>#include<cmath>#include<cstdio>using namespace std;#define manx 100009int x[manx],tree[20][manx];int sum_left[20][manx];__int64 lsum[20][manx];__int64 dp[manx]; void make(int level,int left,int right){    if(left==right) return ;     int mid = (left+right)>>1;    int num = mid - left + 1;    for(int i=left;i<=right;i++){        if(tree[level][i]<x[mid])        num--;    }    int lpos = left, rpos = mid+1;    for(int i=left; i<=right; i++){        if(i==left){            lsum[level][i]=0;//// 初始化             sum_left[level][i]=0;        }            else {            lsum[level][i]=lsum[level][i-1];////初始化             sum_left[level][i]=sum_left[level][i-1];        }            if(tree[level][i]<x[mid]){            sum_left[level][i]++;            lsum[level][i] += tree[level][i];///放在左邊的全部加上             tree[level+1][lpos++] = tree[level][i];            }        else if(tree[level][i]>x[mid])            tree[level+1][rpos++] = tree[level][i];        else {            if(num > 0){                num--;                sum_left[level][i]++;                lsum[level][i] += tree[level][i];///出現與中位元相等的放在左邊的也要加上                 tree[level+1][lpos++]=tree[level][i];            }            else tree[level+1][rpos++]=tree[level][i];         }    }    make(level+1,left,mid);    make(level+1,mid+1,right);}__int64 numl,suml; // 比中位元小的個數 和 總和 __int64 query(int level,int left,int right,int l,int r,int k){    if(l==r) return tree[level][l];    int s, ss, mid = (left+right)>>1;    __int64 temp=0;    if(left==l){        s = 0;        ss = sum_left[level][r];        temp = lsum[level][r]; ///     }    else {        temp = lsum[level][r]-lsum[level][l-1]; /// 求一小段的值         s = sum_left[level][l-1];        ss = sum_left[level][r]-s;    }    int newl,newr;    if(k<=ss){        newl = left + s;        newr = left + s + ss - 1;        return query(level+1,left,mid,newl,newr,k);    }    else {        numl += ss;        suml += temp; ///加上某段求得和         newl = mid + (l-left-s+1);        newr = mid + (r-left-s-ss+1);        return query(level+1,mid+1,right,newl,newr,k-ss);    }} int main(){    int n,q,ca=0,t;    scanf("%d",&t);    while(t--){    scanf("%d",&n);        dp[0]=0;        for(int i=1;i<=n;i++){            scanf("%d",&tree[0][i]);            x[i] = tree[0][i];            dp[i] = x[i] + dp[i-1];        }        sort(x+1,x+n+1);        make(0,1,n);        scanf("%d",&q);        printf("Case #%d:\n",++ca);        while(q--){            int l,r,k;            scanf("%d%d",&l,&r);            l++, r++;            k = (r-l)/2 + 1;            numl = suml = 0;            __int64 val = query(0,1,n,l,r,k);            __int64 sum = val*numl - suml;            numl = r-l+1-numl;            suml = dp[r]-dp[l-1]-suml;            sum += suml - numl*val;            printf("%I64d\n",sum);        }        printf("\n");    }}