POJ 3468 A Simple Problem with Integers 线段树区间更新

#include<stdio.h>
#include<string>
#include<cstring>
#include<queue>
#include<algorithm>
#include<functional>
#include<vector>
#include<iomanip>
#include<math.h>
#include<iostream>
#include<sstream>
#include<set>
#include<climits>
#include<map>
#include<bitset>
using namespace std;
const int MAX=100505;
long long F[6*MAX],Mark[6*MAX]= {0},A[MAX];
void build(int x,int left,int right)
{
if (left==right)
{
F[x]=A[left];
return;
}
int mid=(left+right)/2;
build(x*2,left,mid);
build(x*2+1,mid+1,right);
F[x]=F[x*2]+F[x*2+1];
}
void change(int x,int left,int right,int L,int R,int num)
{
if (left>R||right<L)
return;
if (left>=L&&right<=R)
{
F[x]+=(right-left+1)*num;
Mark[x]+=num;
return;
}
int mid=(left+right)/2;
if (Mark[x])
{
Mark[x*2]+=Mark[x];
Mark[x*2+1]+=Mark[x];
F[x*2]+=(mid-left+1)*Mark[x];
F[x*2+1]+=(right-mid)*Mark[x];
Mark[x]=0;
}
if (L<=mid)
change(x*2,left,mid,L,R,num);
if (R>mid)
change(x*2+1,mid+1,right,L,R,num);
F[x]=F[x*2]+F[x*2+1];
}
long long query(int x,int left,int right,int L,int R)
{
if (left>R||right<L)
return 0;
if (left>=L&&right<=R)
return F[x];
int mid=(left+right)/2;
if (Mark[x])
{
Mark[x*2]+=Mark[x];
Mark[x*2+1]+=Mark[x];
F[x*2]+=(mid-left+1)*Mark[x];
F[x*2+1]+=(right-mid)*Mark[x];
Mark[x]=0;
}
long long Ans=0;
if (L<=mid)
Ans+=query(x*2,left,mid,L,R);
if (R>mid)
Ans+=query(x*2+1,mid+1,right,L,R);
return Ans;
}
int main()
{
int N,Q,a,b,z;
char ch[20];
scanf("%d%d",&N,&Q);
for (int i=1; i<=N; i++)
scanf("%I64d",&A[i]);
memset(Mark,0,sizeof(Mark));
build(1,1,N);
for (int j=1; j<=Q; j++)
{
scanf("%s",ch);
if (ch[0]=='C')
{
scanf("%d%d%d",&a,&b,&z);
change(1,1,N,a,b,z);
}
if (ch[0]=='Q')
{
scanf("%d%d",&a,&b);
printf("%I64d\n",query(1,1,N,a,b));
}
}
return 0;
}