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.
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
这题是简单的插空问题，只要维护每条线段还剩多少空就行，坑点是要用__int64，wa了两次。。
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
__int64 sum;
struct node{
int l,r,num;
}b[8*300000];
void build(int l,int r,int i)
{
int mid;
b[i].l=l;b[i].r=r;b[i].num=r-l+1;
if(l==r)return;
mid=(l+r)/2;
build(l,mid,i*2);
build(mid+1,r,i*2+1);
}
void question(int index,int i)
{
int mid;
if(b[i].l==b[i].r){
sum+=b[i].l;
b[i].num=0;return;
}
if(b[i*2].num>=index)question(index,i*2);
else question(index-b[i*2].num,i*2+1);
b[i].num=b[i*2].num+b[i*2+1].num;
}
int main()
{
int n,m,i,j,T,num1=0,c;
scanf("%d",&T);
while(T--)
{
num1++;
//printf("\n",num1);
scanf("%d%d",&n,&m);
build(1,n,1);
sum=0;
for(i=1;i<=m;i++){
scanf("%d",&c);
question(c,1);
}
printf("Case %d: %I64d\n",num1,sum);
}
return 0;
}