A Simple Problem with Integers(线段树区间更新复习,lazy数组的应用)-------------------蓝桥备战系列
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.
代码:
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<vector>
#include<cmath>
const int maxn=1e5+5;
typedef long long ll;
using namespace std;
struct node
{
ll l,r,sum;
}tree[maxn<<2];
ll lazy[maxn<<2];
void pushup(int m)
{
tree[m].sum=tree[m<<1].sum+tree[m<<1|1].sum;
}
void pushdown(int m,int l)
{
if(lazy[m]!=0)
{
lazy[m<<1]+=lazy[m];
lazy[m<<1|1]+=lazy[m];
tree[m<<1].sum+=lazy[m]*(l-(l>>1));
tree[m<<1|1].sum+=lazy[m]*(l>>1);
lazy[m]=0;
}
}
void build(int m,int l,int r)
{
tree[m].l=l;
tree[m].r=r;
if(l==r)
{
scanf("%lld",&tree[m].sum);
return;
}
int mid=(l+r)>>1;
build(m<<1,l,mid);
build(m<<1|1,mid+1,r);
pushup(m);
}
void update(int m,int l,int r,int val)
{
if(tree[m].l==l&&tree[m].r==r)
{
lazy[m]+=val;
tree[m].sum+=(ll)val*(r-l+1);
return;
}
if(tree[m].l==tree[m].r)
return;
int mid=(tree[m].l+tree[m].r)>>1;
pushdown(m,tree[m].r-tree[m].l+1);
if(r<=mid)
{
update(m<<1,l,r,val);
}
else if(l>mid)
{
update(m<<1|1,l,r,val);
}
else
{
update(m<<1,l,mid,val);
update(m<<1|1,mid+1,r,val);
}
pushup(m);
}
ll query(int m,int l,int r)
{
if(tree[m].l==l&&tree[m].r==r)
{
return tree[m].sum;
}
pushdown(m,tree[m].r-tree[m].l+1);
int mid=(tree[m].l+tree[m].r)>>1;
ll res=0;
if(r<=mid)
{
res+=query(m<<1,l,r);
}
else if(l>mid)
{
res+=query(m<<1|1,l,r);
}
else
{
res+=(query(m<<1,l,mid)+query(m<<1|1,mid+1,r));
}
return res;
}
int main()
{
std::ios::sync_with_stdio(false);
std::cin.tie(0);
int n,m;
cin>>n>>m;
build(1,1,n);
char op[2];
int l,r,val;
for(int t=0;t<m;t++)
{
scanf("%s",op);
if(op[0]=='Q')
{
scanf("%d%d",&l,&r);
printf("%lld\n",query(1,l,r));
}
else
{
scanf("%d%d%d",&l,&r,&val);
update(1,l,r,val);
}
}
return 0;
}
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