poj 3468(简单线段树区间更新)
A Simple Problem with Integers
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 61936 | Accepted: 18934 | |
| 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
#include
#include
#include
using namespace std;
int n,q;
#define ll long long
struct Node{
int left,right;
ll sum,p; //p表示left 到 right 的增量
}a[400010];
void built(int cur,int x,int y){
a[cur].left=x;
a[cur].right=y;
a[cur].p=0;
if(x==y){
scanf("%I64d",&a[cur].sum);
return ;
}
int mid=(x+y)>>1;
built(cur<<1,x,mid);
built(cur<<1|1,mid+1,y);
a[cur].sum=a[cur<<1].sum+a[cur<<1|1].sum;
}
void update(int cur,int x,int y,ll val){
if(a[cur].left==x && a[cur].right==y){
a[cur].sum+=val*(y-x+1);
a[cur].p+=val;
return ;
}
if(a[cur].p){
a[cur<<1].p+=a[cur].p;
a[cur<<1|1].p+=a[cur].p;
a[cur<<1].sum+=a[cur].p*(a[cur<<1].right-a[cur<<1].left+1);
a[cur<<1|1].sum+=a[cur].p*(a[cur<<1|1].right-a[cur<<1|1].left+1);
a[cur].p=0;
}
int mid=(a[cur].left+a[cur].right)>>1;
if(y<=mid)
update(cur<<1,x,y,val);
else if(x>mid)
update(cur<<1|1,x,y,val);
else{
update(cur<<1,x,mid,val);
update(cur<<1|1,mid+1,y,val);
}
a[cur].sum=a[cur<<1].sum+a[cur<<1|1].sum;
}
ll query(int cur,int x,int y){
if(a[cur].left==x && a[cur].right==y){
return a[cur].sum;
}
if(a[cur].p){
a[cur<<1].p+=a[cur].p;
a[cur<<1|1].p+=a[cur].p;
a[cur<<1].sum+=a[cur].p*(a[cur<<1].right-a[cur<<1].left+1);
a[cur<<1|1].sum+=a[cur].p*(a[cur<<1|1].right-a[cur<<1|1].left+1);
a[cur].p=0;
}
int mid=(a[cur].left+a[cur].right)>>1;
if(y<=mid)
return query(cur<<1,x,y);
else if(x>mid)
return query(cur<<1|1,x,y);
else{
return query(cur<<1,x,mid)+query(cur<<1|1,mid+1,y);
}
}
int main(){
while(scanf("%d%d",&n,&q)!=EOF){
built(1,1,n);
while(q--){
char s[2]; scanf("%s",s);
if(s[0]=='C'){
int x,y;
ll val;
scanf("%d%d%I64d",&x,&y,&val);
update(1,x,y,val);
}
else{
int x,y;
scanf("%d%d",&x,&y);
printf("%I64d\n",query(1,x,y));
}
}
}
return 0;
}