HDU 4288 线段树+离散化

时间:2023-03-08 21:07:51

题意:

n个操作

在[1, 100000]  的区间上add 或del数( 必不会重复添加或删除不存在的数)

sum 求出整个集合中 (下标%5 == 3 位置) 的数   的和

注意数据类型要64位

#include <stdio.h>
#include <string.h>
#include <queue>
#include <set>
#include <functional>
#include <map> #define N 101000
#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define Mid(x,y) ((x+y)>>1)
#define ll __int64
using namespace std;
inline ll Max(ll a, ll b){ return a>b?a:b;}
inline ll Min(ll a, ll b){ return a<b?a:b;} ll Point[N]; struct node{
int l,r;
ll sum[5];
int num;
}tree[N*4]; void build( int l, int r, int id){
tree[id].l = l, tree[id].r = r;
memset(tree[id].sum,0,sizeof(tree[id].sum));
tree[id].num = 0; if(l==r)return ;
int mid = Mid(l, r);
build(l, mid, L(id));
build(mid+1, r, R(id));
}
void Updata_up(int id){
tree[id].num = tree[L(id)].num + tree[R(id)].num ; for(int i=0;i<5;i++)
tree[id].sum[i] = tree[L(id)].sum[i]; for(int i=0;i<5;i++) tree[id].sum[ (tree[L(id)].num + i)%5 ] += tree[R(id)].sum[i]; }
void insert(int pos, int id, ll data, bool add){ // add = true 插入data =false 删除data
if(tree[id].l == tree[id].r){
if(add)
{ tree[id].num = 1; tree[id].sum[1] = data;}
else
{ tree[id].num = 0; tree[id].sum[1] = 0; } return ;
} int mid = Mid(tree[id].l, tree[id].r); if( pos <= mid) insert(pos, L(id), data, add);
else insert(pos, R(id), data, add); Updata_up(id);
} struct QUE{
char c;
ll u;
}que[N];
set<ll> tempset; map<ll, int> mymap; void Input(int n){
ll u; char s[5];
tempset.clear();
mymap.clear();
for(int i = 1; i <= n; i++){
scanf("%s", s);
que[i].c = s[0]; if(s[0]!='s')scanf("%I64d",&u), que[i].u = u, tempset.insert(u);
} set<ll> ::iterator p = tempset.begin();
int size = tempset.size();
for(int i = 1; i <= size ; i++,p++){
mymap.insert(pair<ll, int>(*p, i));
Point[i] = *p;
}
}
int go(ll x){
return mymap.find(x) -> second;
}
int main(){
int n;
while(~scanf("%d",&n)){
build(1,100000,1); Input(n); for(int i = 1; i<=n; i++){
ll u = que[i].u;
if(que[i].c == 'a')
insert(go(u),1,u,1); else if(que[i].c == 'd')
insert(go(u),1,u,0); else if(que[i].c == 's')
printf("%I64d\n", tree[1].sum[3]); }
}
return 0;
}
/*
9
add 1
add 2
add 3
add 4
add 5
sum
add 6
del 3
sum ans:
3
4 6
add 1
add 3
add 5
add 7
add 9
sum ans:
5 */