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
POJ Monthly--2007.11.25, Yang Yi
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。。。。。。