Data Structure?
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Problem Description
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.
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.
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
3 2
1 1
10 3
3 9 1
Sample Output
Case 1: 3
Case 2: 14
Case 2: 14
Author
iSea@WHU
Source
思路：taobanzi；
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
const int N=3e5+,M=4e6+,inf=1e9+,mod=1e9+;
const ll INF=1e18+;
int tree[N],n,k;
int lowbit(int x)
{
return x&-x;
}
void update(int x,int change)
{
while(x<=n)
{
tree[x]+=change;
x+=lowbit(x);
}
}
int k_thfind(int K)//树状数组求第K小
{
int sum=;
for(int i=;i>=;i--)
{
if(sum+(<<i)<=n&&tree[sum+(<<i)]<K)
{
K-=tree[sum+(<<i)];
sum+=<<i;
}
}
return sum+;
}
int main(){
int T,cas=;
scanf("%d",&T);
while(T--)
{
memset(tree,,sizeof(tree));
scanf("%d%d",&n,&k);
for(int i=;i<=n;i++)
update(i,);
ll ans=;
for(int i=;i<k;i++)
{
int z;
scanf("%d",&z);
int v=k_thfind(z);
ans+=v;
update(v,-);
}
printf("Case %d: %lld\n",cas++,ans);
}
return ;
}