| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 83822 | Accepted: 25942 | |
| 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
Source
题解：线段树区间更新模版，由于mid比较那块出问题了，错了好半天。。。还有就是注意左边的要是x-x>>1，因为左边可能多一
代码：
#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define SL(x) scanf("%lld",&x)
#define PI(x) printf("%d",x)
#define P_ printf(" ")
#define PL(x) printf("%lld",x)
typedef long long LL;
const int INF=0x3f3f3f3f;
#define ll root<<1
#define rr root<<1|1
#define lson ll,l,mid
#define rson rr,mid+1,r
#define S(x) tree[x].sum
#define L(x) tree[x].lazy
const int MAXN=100010;
LL ans=0;
struct Node{
LL lazy,sum;
};
Node tree[MAXN<<2];
void pushup(int root){
S(root)=S(ll)+S(rr);
}
void pushdown(int root,int x){
if(L(root)){
L(ll)+=L(root);
L(rr)+=L(root);
S(ll)+=L(root)*(x-(x>>1));//注意
S(rr)+=L(root)*(x>>1);
L(root)=0;
}
}
void build(int root,int l,int r){
int mid=(l+r)>>1;
L(root)=0;
if(l==r){
SL(S(root));
//printf("%d %d %lld\n",l,r,S(root));
return;
}
build(lson);
build(rson);
pushup(root);
}
void update(int root,int l,int r,int A,int B,int v){
int mid=(l+r)>>1;
if(l>=A&&r<=B){
L(root)+=v;//注意是+=
S(root)+=v*(r-l+1);
return;
}
pushdown(root,r-l+1);
if(mid>=A)update(lson,A,B,v);//注意
if(mid<B)update(rson,A,B,v);//注意
pushup(root);
}
void query(int root,int l,int r,int A,int B){
int mid=(l+r)>>1;
//printf("%d %d %lld\n",l,r,S(root));
// PI(A);P_;PI(B);
if(l>=A&&r<=B){
ans+=S(root);
//printf("%d %d %lld\n",l,r,S(root));
return;
}
pushdown(root,r-l+1);
if(mid>=A)query(lson,A,B);//
if(mid<B)query(rson,A,B);//
}
int main(){
int N,Q;
char s[2];
int A,B,v;
while(~scanf("%d%d",&N,&Q)){
build(1,1,N);
while(Q--){
scanf("%s",s);
if(s[0]==‘Q‘){
scanf("%d%d",&A,&B);
ans=0;
query(1,1,N,A,B);
printf("%lld\n",ans);
}
else{
scanf("%d%d%d",&A,&B,&v);
update(1,1,N,A,B,v);
}
}
}
return 0;
}
A Simple Problem with Integers(线段树，区间更新)
原文：http://www.cnblogs.com/handsomecui/p/5125668.html