Crawling in process...Crawling failedTime
Limit:5000MS Memory Limit:131072KB
64bit IO Format:%I64d & %I64u
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
这道题是我最痛苦的一次，思路很快出来，敲的也快，可惜却因为敲少了个+号，少了个和，少了个long long ，却是绝大部分数据都OK，花了一个下午的狂调
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#define N 100300
using namespace std;
class node
{
public:
int l,r;
long long sum,lazy;
}root[4*N];
long long shu[N];
void build(int t,int l,int r)
{
root[t].l=l;root[t].r=r;root[t].lazy=0;
if(root[t].l==root[t].r){cin>>root[t].sum;return;}
build(t<<1,l,(l+r)/2);build(t<<1|1,(l+r)/2+1,r);
root[t].sum=root[t<<1].sum+root[t<<1|1].sum;
}
void update(int t,int l,int r,long long date)
{
if(root[t].l==l&&root[t].r==r)
{root[t].sum+=(root[t].r-root[t].l+1)*date;root[t].lazy+=date;return;}//这里的+原下没加WA
else
{
if(root[t].lazy!=0)
{
root[t<<1].lazy+=root[t].lazy;root[t<<1|1].lazy+=root[t].lazy;
root[t<<1].sum+=(root[t<<1].r-root[t<<1].l+1)*root[t].lazy;
root[t<<1|1].sum+=(root[t<<1|1].r-root[t<<1|1].l+1)*root[t].lazy;
root[t].lazy=0;
}
if(root[t<<1].r>=r)update(t<<1,l,r,date);
else if(root[t<<1|1].l<=l)update(t<<1|1,l,r,date);
else {update(t<<1,l,root[t<<1].r,date);update(t<<1|1,root[t<<1|1].l,r,date);}
root[t].sum=root[t<<1].sum+root[t<<1|1].sum;//这个很关键没加WA
}
}
long long query(int t,int l,int r)
{
long long s;
if(root[t].l==l&&root[t].r==r)return root[t].sum;
else
{
if(root[t].lazy!=0)
{
root[t<<1].lazy+=root[t].lazy;root[t<<1|1].lazy+=root[t].lazy;
root[t<<1].sum+=(root[t<<1].r-root[t<<1].l+1)*root[t].lazy;
root[t<<1|1].sum+=(root[t<<1|1].r-root[t<<1|1].l+1)*root[t].lazy;
root[t].lazy=0;
}
if(root[t<<1].r>=r)s=query(t<<1,l,r);
else if(root[t<<1|1].l<=l)
s=query(t<<1|1,l,r);
else s=query(t<<1,l,root[t<<1].r)+query(t<<1|1,root[t<<1|1].l,r);
}
return s;
}
int main()
{
int n,m;
scanf("%d%d",&n,&m);
build(1,1,n);
char f;
long long a,b,c;
for(int i=0;i<m;i++)
{
cin>>f>>a>>b;
if(f=='C')
{
cin>>c;
update(1,a,b,c);
}
else
cout<<query(1,a,b)<<"\n";
}
}