HDU4893 Wow! Such Sequence! 多校练习赛 树状数组

三个操作

1、单点更新

2、[l, r] 区间自动修改成最近的 Fibonacci 数据

3、[l,r] 区间求和

= =其实。。 自动更新的数据是很少的。。 只用把它存在 set里面。。用一个删一个就结束了。。

用树状数组维护。。

//tpl
//ipqhjjybj_tpl.h
//header.h
#include <cstdio>
#include <cstdlib>
#include <map>
#include <set>
#include <algorithm>
#include <cstring>
#include <iostream>
#include <vector>
#include <string>
#include <queue>

#define mp(x,y) make_pair(x,y)
#define pii pair<int,int>
#define pLL pair<long long ,long long>
#define rep(i,j,k) for(int i = j; i < k;i++)
#define repn(i,j,k) for(int i = j; i <= k;i++)

using namespace std;
#define lowbit(t) (t&(-t))
const int N = 100005;
long long F[N];
long long sum[N],a[N],c[N];
int n,m;
void update(int pos,long long val){
	for(; pos <= n; pos += lowbit(pos))
		sum[pos] += val;
}
long long QUERY(int pos){
	long long s = 0;
	for( ; pos ; pos -= lowbit(pos)){
		s += sum[pos];
		//printf("pos=%d\n",pos);
	}
	return s;
}
int ts = 1;
long long fuckIt(int i){
	long long val = a[i];
	int pp = lower_bound(F+1,F+ts,val) - F;
	//printf("pp=%d\n",pp);
	//printf("a[%d]=%lld F[%d]=%lld F[%d]=%lld\n",i,a[i],pp-1,F[pp-1],pp,F[pp]);
	if(val == F[pp]){ }
	else if( val - F[pp-1] > F[pp]-val){
		val = F[pp]-val;
		val = a[i] + val;
	}else{
		val = F[pp-1]-val;
		val = a[i] + val;
	}
	return val;
}
set<int> fs;
set<int> ks;
int main(){
	F[0] = F[1] = 1;
	do{
		ts++;
		F[ts] = F[ts-1]+F[ts-2];
	}while(F[ts] > 1);
	int u,v,w;
	while(scanf("%d %d",&n,&m)!=EOF){
		
		for(int i=1;i<=n;i++) sum[i] = 0;
		for(int i=1;i<=n;i++) c[i] = 1 , a[i]=0;
		fs.clear();
		for(int i=1;i<=n;i++) fs.insert(i);
		ks.clear();
		while(m--){
			scanf("%d %d %d",&u,&v,&w);
			if(u == 1){
				update(v,w);
				a[v]+=w;
				c[v]=fuckIt(v);
				fs.insert(v);
			}else if(u==3){
				for(set<int>::iterator it = fs.lower_bound(v); it !=fs.end(); it++){
					int i = *it;
					if(c[i]!=0 &&  i <= w){
						if( a[i] != c[i]) update(i,c[i]-a[i]);
						a[i] = c[i];
						c[i] = 0;
						ks.insert(i);
					}
					if(i > w) break;
				}
				for(set<int>::iterator it =ks.begin(); it!=ks.end(); it++ ){
					fs.erase(*it);
				}
				ks.clear();

			}else if(u==2){
				long long ans = QUERY(w)-QUERY(v-1);

				cout << ans <<endl;
			}
		}
	}
	return 0;
}