| A Simple Problem with Integers | ||||||
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| 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 | ||||||
|
Input contain multiple test cases,for each case: The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000. |
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| Output | ||||||
| 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. | ||||||
| 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 |
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| Sample Output | ||||||
4 55 9 15 |
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| Source | ||||||
| POJ Monthly--2007.11.25, Yang Yi | ||||||
| Recommend | ||||||
| `Wind @Hrbust |
线段树区间更新，如果真的全部更新每个节点，会很耗费时间，所以在可行的节点上标记一下，这样下次查询的时候如果查询到这个节点，就往下更新一次
#include<cstdio>
#include<cmath>
#include<queue>
#include<iostream>
using namespace std;
const int m=100000;
typedef long long LL;
struct node
{
LL Inc;
LL sum;
}a[m<<2];
LL b[m+1];
void build(int L,int R,int rt)
{
a[rt].Inc=0;
if(L==R)
{
// scanf("%lld",&b[L]);
a[rt].sum=b[L];
return ;
}
int lr=rt<<1,rr=rt<<1|1;
int mid=(L+R)>>1;
build(L,mid,lr);
build(mid+1,R,rr);
a[rt].sum=a[lr].sum+a[rr].sum;
}
void update(int l,int r,int L,int R,int rt,LL x)
{
if(L==l&&R==r)
{
a[rt].Inc+=x;
return ;
}
a[rt].sum+=(r-l+1)*x;
int m=(L+R)>>1;
int lr=rt<<1,rr=rt<<1|1;
if(r<=m)
{
update(l,r,L,m,lr,x);
}
else if(l>m)
{
update(l,r,m+1,R,rr,x);
}
else
{
update(l,m,L,m,lr,x);
update(m+1,r,m+1,R,rr,x);
}
}
LL query(int l,int r,int L,int R,int rt)
{
if(l<=L&&r>=R)
{
return a[rt].sum+a[rt].Inc*(r-l+1);
}
a[rt].sum+=(R-L+1)*a[rt].Inc;
int m=(L+R)>>1;
int lr=rt<<1,rr=rt<<1|1;
if(a[rt].Inc)
{
//update(L,m,L,m,lr,a[rt].Inc);
// update(m+1,R,m+1,R,rr,a[rt].Inc);
a[lr].Inc+=a[rt].Inc;
a[rr].Inc+=a[rt].Inc;
}
a[rt].Inc=0;
if(r<=m)
{
return query(l,r,L,m,lr);
}
else if(l>m)
{
return query(l,r,m+1,R,rr);
}
LL left=query(l,m,L,m,lr);
LL right=query(m+1,r,m+1,R,rr);
return left+right;
}
int main()
{
int N,Q;
while(~scanf("%d%d",&N,&Q))
{
int i;
for(i=1;i<=N;i++)
{
scanf("%lld",&b[i]);
}
build(1,N,1);
char c;
int l,r;
LL x;
while(Q--)
{
cin>>c;
if(c==‘Q‘)
{
scanf("%d%d",&l,&r);
printf("%lld\n",query(l,r,1,N,1));
}
else if(c==‘C‘)
{
// cout<<"dsfdsf"<<endl;
scanf("%d%d%lld",&l,&r,&x);
update(l,r,1,N,1,x);
}
}
}
return 0;
}
时间： 2024-08-10 23:30:30