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.
Technical Specification
1. 1 <= T <= 128
2. 1 <= K <= N <= 262 144
3. 1 <= Ki <= N - i + 1
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
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+10,M=4e6+10,inf=1e9+10,mod=1e9+7; const ll INF=1e18+10; 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=0; for(int i=18;i>=0;i--) { if(sum+(1<<i)<=n&&tree[sum+(1<<i)]<K) { K-=tree[sum+(1<<i)]; sum+=1<<i; } } return sum+1; } int main(){ int T,cas=1; scanf("%d",&T); while(T--) { memset(tree,0,sizeof(tree)); scanf("%d%d",&n,&k); for(int i=1;i<=n;i++) update(i,1); ll ans=0; for(int i=0;i<k;i++) { int z; scanf("%d",&z); int v=k_thfind(z); ans+=v; update(v,-1); } printf("Case %d: %lld ",cas++,ans); } return 0; }