Data structure is one of the basic skills for Computer Science students, which is a particular way of storing and organizing data in a computer so that it can be used efficiently. Today let me introduce a data-structure-like problem for you.
Original, there are N numbers, namely 1, 2, 3...N. Each round, iSea find out the Ki-th smallest number and take it away, your task is reporting him the total sum of the numbers he has taken away.
Input
The first line contains a single integer T, indicating the number of test cases.
Each test case includes two integers N, K, K indicates the round numbers. Then a line with K numbers following, indicating in i (1-based) round, iSea take away the Ki-th smallest away.
Technical Specification
1. 1 <= T <= 128
2. 1 <= K <= N <= 262 144
3. 1 <= Ki <= N - i + 1
Output
For each test case, output the case number first, then the sum.
Sample Input
2 3 2 1 1 10 3 3 9 1
Sample Output
Case 1: 3 Case 2: 14
题意：1到n n个数，一共有 k 次操作，每次取出第 ki 小的数。问所有取出数字之和。
题解：线段树维护下区间还有多少数，更新到节点即可
#include<iostream>
#include<cstdio>
using namespace std;
const int N=270000;
typedef long long ll;
struct node{
int l,r,sum;
}tree[N<<2];
int n,k;
void build(int l,int r,int cur)
{
tree[cur].l=l;
tree[cur].r=r;
tree[cur].sum=r-l+1;
if(l==r)return;
int mid=(r+l)>>1;
build(l,mid,cur<<1);
build(mid+1,r,cur<<1|1);
}
int query(int cur,int pos)
{
if(tree[cur].l==tree[cur].r)
{
tree[cur].sum=0;
return tree[cur].l;
}
int res;
if(tree[cur<<1].sum>=pos) res=query(cur<<1,pos);
else
{
res=query(cur<<1|1,pos-tree[cur<<1].sum);
}
tree[cur].sum=tree[cur<<1].sum+tree[cur<<1|1].sum;
return res;
}
int main()
{
int T,nn=1;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&k);
build(1,n,1);
ll ans=0;
int x;
for(int i=1;i<=k;i++)
{
scanf("%d",&x);
ans+=query(1,x);
}
printf("Case %d: %lld\n",nn++,ans);
}
return 0;
}