Hdu 4288 Coder(从小到大排列的集合中下标模5为3的数的和)

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分类专栏： codeforces

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Coder

Problem Description

　　In mathematics and computer science, an algorithm describes a set of procedures or instructions that define a procedure. The term has become increasing popular since the advent of cheap and reliable computers. Many companies now employ a single coder to write an algorithm that will replace many other employees. An added benefit to the employer is that the coder will also become redundant once their work is done. ¹
　　You are now the signle coder, and have been assigned a new task writing code, since your boss would like to replace many other employees (and you when you become redundant once your task is complete).
Your code should be able to complete a task to replace these employees who do nothing all day but eating: make the digest sum.
　　By saying “digest sum” we study some properties of data. For the sake of simplicity, our data is a set of integers. Your code should give response to following operations:
　　1. add x – add the element x to the set;
　　2. del x – remove the element x from the set;
　　3. sum – find the digest sum of the set. The digest sum should be understood by

　　where the set S is written as {a₁, a₂, ... , a_k} satisfying a₁ < a₂ < a₃ < ... < a_k
　　Can you complete this task (and be then fired)?
------------------------------------------------------------------------------
¹ See http://uncyclopedia.wikia.com/wiki/Algorithm

Input

　　There’re several test cases.
　　In each test case, the first line contains one integer N ( 1 <= N <= 10⁵ ), the number of operations to process.
　　Then following is n lines, each one containing one of three operations: “add x” or “del x” or “sum”.
　　You may assume that 1 <= x <= 10⁹.
　　Please see the sample for detailed format.
　　For any “add x” it is guaranteed that x is not currently in the set just before this operation.
　　For any “del x” it is guaranteed that x must currently be in the set just before this operation.
　　Please process until EOF (End Of File).

Output

　　For each operation “sum” please print one line containing exactly one integer denoting the digest sum of the current set. Print 0 if the set is empty.

Sample Input

  
  
   
   9
add 1
add 2
add 3
add 4
add 5
sum
add 6
del 3
sum
6
add 1
add 3
add 5
add 7
add 9
sum

Sample Output

  
  
   
   3
4
5

   
   
    
    
     
     Hint
    
    
    
    C++ maybe run faster than G++ in this problem.

Source

2012 ACM/ICPC Asia Regional Chengdu Online

题意: 有三种类型的操作，（ 1 ） ."add x" ，表示往集合里添加数 x 。（ 2 ） . “ del x ”表示将集合中数 x 删除。（ 3 ） . “ sum ”求出从小到大排列的集合中下标模 5 为 3 的数的和。集合中的数都是唯一的。 the set S is written as {a1, a2, ... , ak} satisfying a1 < a2 < a3 < ... < ak

(1<=n<=1e5,1<=x<=1e9)

思路:因为插入的数不一定是有序的,所以我们要先对其离散化,然后每个位置维护一下余数分别为0,1,2,3,4的数的和还有这个区间总共出现了多少数,然后简单进行更新就可以了。

#include<bits/stdc++.h>
using namespace std;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
const int maxn=100100;
typedef long long ll;
ll sumv[maxn*4][5],cnt[maxn*4],num[maxn*4];
int t[maxn],pos,v;
struct q{
    int op,num;
}Q[maxn];

void build(int l,int r,int rt){
    for(int i=0;i<5;i++)
        sumv[rt][i]=0;
    cnt[rt]=0;
    if(l==r){
        num[rt]=t[l];
        return ;
    }
    int mid=(l+r)>>1;
    build(lson);
    build(rson);
}

void pushup(int rt){
    cnt[rt]=cnt[rt<<1]+cnt[rt<<1|1];
    cnt[rt]%=5;
    for(int i=0;i<5;i++)
        sumv[rt][i]=sumv[rt<<1][i]+sumv[rt<<1|1][(i+5-cnt[rt<<1])%5];
}

void update(int l,int r,int rt){
    if(l==r){
        sumv[rt][1]+=num[rt]*v;
        cnt[rt]+=v;
        return ;
    }
    int mid=(l+r)>>1;
    if(pos<=mid)
        update(lson);
    else
        update(rson);
    pushup(rt);
}

int main(){
    int m;
    while(scanf("%d",&m)!=EOF){
        char op[10];
        int n=0;
        for(int i=1;i<=m;i++){
            scanf("%s",op);
            if(op[0]=='a'){
                Q[i].op=1;
                scanf("%d",&Q[i].num);
                t[++n]=Q[i].num;
            }
            else if(op[0]=='s')
                Q[i].op=3;
            else
                scanf("%d",&Q[i].num),Q[i].op=2;
        }
        sort(t+1,t+n+1);
        n=unique(t+1,t+n+1)-t-1;
        build(1,n,1);
        for(int i=1;i<=m;i++){
            if(Q[i].op==1)
                v=1,pos=lower_bound(t+1,t+n+1,Q[i].num)-t,update(1,n,1);
            else if(Q[i].op==2)
                v=-1,pos=lower_bound(t+1,t+n+1,Q[i].num)-t,update(1,n,1);
            else
                printf("%lld\n",sumv[1][3]);
        }
    }
    return 0;
}