G - A Simple Problem with Integers POJ - 3468

最新推荐文章于 2022-08-09 17:16:44 发布

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线段树&&树状数组

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You have N integers, A₁,A₂, ... , A_N. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.

Input

The first line contains two numbers N andQ. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A₁,A₂, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a₊₁, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a, A_a₊₁, ... ,A_b.

Output

You need to answer all Q commands in order. One answer in a line.

Sample Input

10 5
1 2 3 4 5 6 7 8 9 10
Q 4 4
Q 1 10
Q 2 4
C 3 6 3
Q 2 4

Sample Output

Hint

The sums may exceed the range of 32-bit integers.

题意：给你n个整数，有两种操作，第一种，当字母为Q时，输出区间【a,b】的和，为C时，将区间【a,b】整体加c;

思路：线段树模板，注意数据范围诶，另外延迟标记用的可以^__^,哈哈哈；

下面附上代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#define ll o<<1
#define rr o<<1|1
using namespace std;
typedef long long LL;
const int M=100005;
struct Tree
{
	LL sum;
	LL lazy;
	LL l,r,len; 
};
Tree tree[M<<2];
void pushup(LL o)
{
	tree[o].sum=tree[ll].sum+tree[rr].sum;
}
void pushdown(LL o)
{
	if(tree[o].lazy)
	{
		tree[ll].lazy+=tree[o].lazy;
		tree[rr].lazy+=tree[o].lazy;
		tree[ll].sum+=tree[o].lazy*tree[ll].len;
		tree[rr].sum+=tree[o].lazy*tree[rr].len;
		tree[o].lazy=0;
	}
}
void build(LL o, LL L,LL R)
{
	tree[o].l=L,tree[o].r=R;
	tree[o].len=R-L+1;
	tree[o].lazy=0;
	if(L==R)
	{
		LL p;
		scanf("%lld",&p);
		tree[o].sum=p;
		return ;
	}
	LL mid=(L+R)>>1;
	build(o<<1,L,mid);
	build(o<<1|1,mid+1,R);
	pushup(o);
}
void update(LL o,LL L,LL R,LL v)
{
	if(tree[o].l>=L&&tree[o].r<=R)
	{
		tree[o].lazy+=v;
		tree[o].sum+=v*tree[o].len;   
		return ;   
	}
	pushdown(o);
	LL mid=(tree[o].l+tree[o].r)>>1;
	if(R<=mid)
		update(ll,L,R,v);
	else if(L>mid)
		update(rr,L,R,v);
	else
	{
		update(ll,L,mid,v);
		update(rr,mid+1,R,v);
	}
	pushup(o);
}
LL Query(LL o,LL L,LL R)
{
	if(tree[o].l>=L&&tree[o].r<=R)
		return tree[o].sum;
	LL mid=(tree[o].l+tree[o].r)>>1;
	pushdown(o);
	if(R<=mid)
		return Query(ll,L,R);
	else if(L>mid)
		return Query(rr,L,R);
	else
		return 	Query(ll,L,mid)+Query(rr,mid+1,R);
}
int main()
{
	LL N,Q1;
	char s[3]={0};
	LL a,b,c;
	scanf("%lld %lld",&N,&Q1);
	build(1,1,N);
	while(Q1--)
	{
		scanf("%s",s);
		if(s[0]=='Q')
		{
			scanf("%lld %lld",&a,&b);
			LL k=Query(1,a,b);
			printf("%lld\n",k);
		}
		else
		{
			scanf("%lld %lld %lld",&a,&b,&c);
			update(1,a,b,c);
		}
	}
	return 0;
}