Poj 3468-A Simple Problem with Integers 线段树,树状数组

时间：2021-10-22 08:58:43
 
A Simple Problem with Integers
Time Limit: 5000MS   Memory Limit: 131072K
Total Submissions: 85851   Accepted: 26685
Case Time Limit: 2000MS

Description

You have N integers, A1, A2, ... , AN. 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 and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A1, A2, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of Aa, Aa+1, ... , Ab.

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

4

55

9

15

Hint

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

Source

 
题意:给一串数字，每次区间加上一个数，或询问区间和。
 
题解:
线段树水过，打个lazy标记就好。。。
线段树程序：
 #include<cstdio>

 #include<cstring>

 #include<cstdlib>

 #include<cmath>

 #include<iostream>

 #include<algorithm>

 using namespace std;

 #define LL long long

 #define MAXN 100010

 struct node

 {

     LL left,right,sum,tag;

 }tree[MAXN*];

 LL A[MAXN];

 LL read()

 {

     LL s=,fh=;char ch=getchar();

     while(ch<''||ch>''){if(ch=='-')fh=-;ch=getchar();}

     while(ch>=''&&ch<=''){s=s*+(ch-'');ch=getchar();}

     return s*fh;

 }

 void Pushup(LL k)

 {

     tree[k].sum=tree[k*].sum+tree[k*+].sum;

 }

 void Pushdown(LL k,LL l,LL r)

 {

     if(tree[k].tag!=)

     {

         tree[k*].tag+=tree[k].tag;

         tree[k*+].tag+=tree[k].tag;

         LL mid=(l+r)/;

         tree[k*].sum+=tree[k].tag*(mid-l+);

         tree[k*+].sum+=tree[k].tag*(r-mid);

         tree[k].tag=;

     }

 }

 void Build(LL k,LL l,LL r)

 {

     tree[k].left=l;tree[k].right=r;

     if(l==r)

     {

         tree[k].sum=A[l];

         return;

     }

     LL mid=(l+r)/;

     Build(k*,l,mid);Build(k*+,mid+,r);

     Pushup(k);

 }

 void Add(LL k,LL ql,LL qr,LL add)

 {

     //Pushdown(k,tree[k].left,tree[k].right);

     if(ql<=tree[k].left&&tree[k].right<=qr){tree[k].tag+=add;tree[k].sum+=(tree[k].right-tree[k].left+)*add;return;}

     Pushdown(k,tree[k].left,tree[k].right);

     LL mid=(tree[k].left+tree[k].right)/;

     if(qr<=mid)Add(k*,ql,qr,add);

     else if(ql>mid)Add(k*+,ql,qr,add);

     else {Add(k*,ql,mid,add);Add(k*+,mid+,qr,add);}

     Pushup(k);

 }

 LL Sum(LL k,LL ql,LL qr)

 {

     //Pushdown(k,tree[k].left,tree[k].right);

     if(ql<=tree[k].left&&tree[k].right<=qr)return tree[k].sum;

     Pushdown(k,tree[k].left,tree[k].right);

     LL mid=(tree[k].left+tree[k].right)/;

     if(qr<=mid)return Sum(k*,ql,qr);

     else if(ql>mid)return Sum(k*+,ql,qr);

     else return Sum(k*,ql,mid)+Sum(k*+,mid+,qr);

 }

 int main()

 {

     LL N,Q,i,a,b,c;

     char fh[];

     N=read();Q=read();

     for(i=;i<=N;i++)A[i]=read();

     Build(,,N);

     for(i=;i<=Q;i++)

     {

         scanf("\n%s",fh);

         if(fh[]=='Q')

         {

             a=read();b=read();

             printf("%lld\n",Sum(,a,b));

         }

         else

         {

             a=read();b=read();c=read();

             Add(,a,b,c);

         }

     }

     return ;

 }

树状数组不会。。。

先附上fhq神犇的题解：

上次NOI集训的时候，一位福建女神牛和我探讨过这题能不能用BIT，我当时

的答复是可以，因为“扩展树状数组”这个东西理论上可以实现一般线段树

可以实现的东西，且空间上的常数好一点。但是对于“扩展树状数组”，这

个东西是我一时兴起想到的玩意，没有进行更多的研究，没查到任何的资料

，更没有想过如何把线段树著名的lazy思想照搬上去，以及动态开内存的解

决方案。

权衡利弊，我想了一个使用两棵BIT的替代方法来解决这题，并且成功地将

内存使用做到了1728K。这恐怕是带标记的扩展树状数组达不到的。

记录两个BIT，

设数列为A[i],BIT1的每个元素B1[i]=A[i]*i,

BIT2的每个元素B2[i]=A[i]

则：sum{A[i]|i<=a}=sum{B1[i]|i<=a}+(sum{B2[i]|1<=i<=N}-sum{B2[i]|i<=a})*a

sum{A[i]|a<=i<=b}=sum{A[i]|i<=b}-sum{A[i]|i<a}

这样就十分巧妙的解决了！

关键代码：

int N;

struct BIT{

	long long a[NMax];

	void ins(int x,long long k){

		for (;x<N;x+=((x+1)&-(x+1)))a[x]+=k;

	}

	long long ask(int x){

		long long ret;

		for (ret=0;x>=0;x-=((x+1)&-(x+1)))ret+=a[x];

		return ret;

	}

}B1,B2;

long long B2S;

void Add(int a,long long x){

	//printf("Add %d %I64d\n",a,x);

	B1.ins(a,x*((long long)a+1));

	B2.ins(a,x);B2S+=x;

}

void Add(int a,int b,long long x){

	Add(b,x);

	if (a)Add(a-1,-x);

}

long long Ask(int a){

	//printf("Ask %d\n",a);

	long long ret=B1.ask(a)+(B2S-B2.ask(a))*(long long)(a+1);

	//printf("=%I64d\n",ret);

	return ret;

}

long long Ask(int a,int b){

	long long ret;

	ret=Ask(b);

	if (a)ret-=Ask(a-1);

	return ret;

}

填坑中。。。。。。

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