C - Data Structure?
Time Limit:5000MS Memory Limit:65536KB 64bit IO Format:%I64d
& %I64u
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4217
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
这道题目能够用线段树，或者是树状数组来解决
对于题意而言，他的意思是给你n个数字，分别为1...n然后是
给你询问，让你去除这个数，可是询问的内容是序列，比方说1,2,3假设我取出了2，那么当询问为2的时候就是3了。由于序列变为了1,3
然后是通过线段树的标记功能
/*
Problem : 4217 ( Data Structure?
) Judge Status : Accepted
RunId : 13881893 Language : C++ Author : 24862486
Timer : 998ms
Code Render Status : Rendered By HDOJ C++ Code Render Version 0.01 Beta
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define lson rt << 1 , L , mid
#define rson rt << 1 | 1 , mid + 1 , R
const int maxn=262144+5;
int n,k,ki,T;
int sum[maxn<<2];
void pushup(int rt) {
sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
void build(int rt,int L,int R) {
if(L==R) {
sum[rt]=1;
return;
}
int mid=(L+R)>>1;
build(lson);
build(rson);
pushup(rt);
}
int query(int p,int rt,int L,int R) {
if(L==R) {
sum[rt]=0;
return R;
}
int mid=(L+R)>>1;
int res;
if(sum[rt<<1]>=p)res=query(p,lson);//向终点靠拢
else res=query(p-sum[rt<<1],rson);
pushup(rt);
return res;
}
int main() {
scanf("%d",&T);
for(int t=1; t<=T; t++) {
scanf("%d%d",&n,&k);
build(1,1,n);
long long res=0;
for(int i=0; i<k; i++) {
scanf("%d",&ki);
res+=query(ki,1,1,n);
}
printf("Case %d: %lld\n",t,res);
}
return 0;
}