POJ_3468_A Simple Problem_线段树

快两点了还不困，明天要死。

题意：

区间更新，加上一个值，区间查询区间和。

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.

线段树的区间更新，但是看Discuss里面大部分都是splay。。。。

书上用的是树状数组的变形。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<cstdlib>
#include<fstream>
using namespace std;
#define mxn 100010
#define mxq 100010
int n,q;
long long a[mxn];
class node{
public:
int ll,rr;
long long sum;
long long flag;
node(){}
};
class segment_tree{
private:
node nd[mxn<<2];
public:
void build(int l,int r,int id){
nd[id].ll=l;
nd[id].rr=r;
nd[id].flag=0;
if(l==r){
nd[id].sum=a[l];
return;
}
int m=(l+r)>>1,ls=id<<1,rs=ls|1;
build(l,m,ls);
build(m+1,r,rs);
nd[id].sum=nd[ls].sum+nd[rs].sum;
}
void down(int);
void update(int l,int r,long long x,int id){
if(l==nd[id].ll&&r==nd[id].rr){
nd[id].flag+=x;
nd[id].sum+=x*(r-l+1);
return;
}
if(nd[id].flag)	down(id);
int m=(nd[id].ll+nd[id].rr)>>1,ls=id<<1,rs=ls|1;
if(r<=m)	update(l,r,x,ls);
else if(l>m)	update(l,r,x,rs);
else{
update(l,m,x,ls);
update(m+1,r,x,rs);
}
nd[id].sum=nd[ls].sum+nd[rs].sum;
}
long long find(int l,int r,int id){
if(nd[id].ll==l&&nd[id].rr==r)
return nd[id].sum;
if(nd[id].flag)	down(id);
int m=(nd[id].ll+nd[id].rr)>>1,ls=id<<1,rs=ls|1;
if(r<=m)	return find(l,r,ls);
else if(l>m)	return find(l,r,rs);
else    return find(l,m,ls)+find(m+1,r,rs);
}
void printAns(int l,int r){
long long ans=find(l,r,1);
printf("%I64d\n",ans);
}
}Tree;
void segment_tree::down(int id){
if(nd[id].ll==nd[id].rr)	return;
int m=(nd[id].ll+nd[id].rr)>>1,ls=id<<1,rs=ls|1;
update(nd[id].ll,m,nd[id].flag,ls);
update(m+1,nd[id].rr,nd[id].flag,rs);
nd[id].flag=0;
}
int main(){
scanf("%d%d",&n,&q);
for(int i=0;i<n;++i)	scanf("%I64d",&a[i]);
Tree.build(0,n-1,1);
int a,b;
char type;
long long c;
for(int i=0;i<q;++i){
getchar();
scanf("%c",&type);
if(type=='Q'){
scanf("%d%d",&a,&b);
Tree.printAns(a-1,b-1);
}
else{
scanf("%d%d%I64d",&a,&b,&c);
Tree.update(a-1,b-1,c,1);
}
}
//system("pause");
return 0;
}