求助loj6281
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  • 楼主EDqwq
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  • 发布时间2021/2/2 20:51
  • 上次更新2023/11/5 03:54:09
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求助loj6281
294562
EDqwq楼主2021/2/2 20:51
#include<bits/stdc++.h>

#define int long long
#define mem(x) memset(x,0,sizeof(x))

using namespace std;

int read(){
   int s = 0,w = 1;
   char ch = getchar();
   while(ch < '0' || ch > '9'){if(ch == '-')w = -1;ch = getchar();}
   while(ch >= '0' && ch <= '9')s = s * 10 + ch - '0',ch = getchar();
   return s * w;
}

int n;
int a[1000010];
int l[1000010],r[1000010];
int block;
int num[1000010];
int lazy[1000010];
int len;
int times[1000010];
int sum[1000010];

void update(int x,int y){
	int posl = num[x];
	int posr = num[y];
	if(posl == posr){
		for(int i = x;i <= y;i ++){
			int vx = sqrt(a[i]);
			a[i] = vx;
			sum[num[posl]] -= (a[i] - vx);
		}
		return ;
	}
	for(int i = x;i <= r[posl];i ++){
		int vx = sqrt(a[i]);
		a[i] = vx;
		sum[num[posl]] -= (a[i] - vx);
	}
	for(int i = num[x] + 1;i <= num[y] - 1;i ++)lazy[i] ++; 
	for(int i = l[posr];i <= y;i ++){
		int vx = sqrt(a[i]);
		a[i] = vx;
		sum[num[posl]] -= (a[i] - vx);
	}
	return ;
}

int query(int x,int y){
	int posl = num[x];
	int posr = num[y];
	int ans = 0;
	if(posl == posr){
		for(int i = x;i <= y;i ++){
			if(a[i] == 0)continue;
			if(lazy[posl] >= 5){
				ans ++;
				continue;
			}
			int s = a[i];
			for(int j = 1;j <= lazy[posl];j ++){
				s = sqrt(s);
			}
			ans += s;
		}
		return ans;
	}
	for(int i = x;i <= r[posl];i ++){ 
		if(a[i] == 0)continue;
		if(lazy[num[i]] >= 5){
			ans ++;
			continue;
		}
		int s = a[i];
		for(int j = 1;j <= lazy[num[i]];j ++){
			s = sqrt(s);
		}
		ans += s;
	}
	for(int i = num[x] + 1;i <= num[y] - 1;i ++){
		if(lazy[i] >= 5)ans += block - times[i];
		else {
			int s = a[i];
			for(int j = 1;j <= lazy[i];j ++){
				s = sqrt(s);
			}
			ans += s;
		}
	}
	for(int i = l[posr];i <= y;i ++){
		if(a[i] == 0)continue;
		if(lazy[num[i]] >= 5){
			ans ++;
			continue;
		}
		int s = a[i];
		for(int j = 1;j <= lazy[num[i]];j ++){
			s = sqrt(s);
		}
		ans += s;
	}
	return ans;
}

signed main(){
	cin>>n;
	block = sqrt(n);
	len = ceil(n * 1.0 / block);
	for(int i = 1;i <= len;i ++){
		l[i] = (i - 1) * block + 1;
		r[i] = i * block;
	}
	for(int i = 1;i <= n;i ++){
		a[i] = read();
		num[i] = (i - 1) / block + 1;
		sum[num[i]] += a[i];
		if(a[i] == 0)times[num[i]] ++;
	}
	r[len] = n;
	for(int i = 1;i <= n;i ++){
		int op;
		int x,y,w;
		op = read(),x = read(),y = read(),w = read();
		if(op == 0){
			update(x,y);
		}
		else {
			printf("%lld\n",query(x,y));
		}
	}
	return 0;
}

虽然不吃电脑了,但是看看代码还是可以的

具体操作就是区间根号,区间求和

2021/2/2 20:51
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