题目链接:  http://poj.org/problem?id=3468

题目大意:  给出N个数，和M次查询

C a b c  区间[a,b]的值都加上c

Q a b     查询区间[a,b]值的和

解题思路:  线段树区间lazy延迟更新，每次插入区间标记lazy

下次再操作此区间时用lazy更新下面的子树

每个结点存储值是区间的和

更新和查询的时间复杂度都是O(logN)

代码:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAX 201000
#define MID(a,b) (a+b)>>1
#define L(a) a<<1
#define R(a) (a<<1|1)
typedef struct{int left,right;long long int sum,add;
}Node;Node Tree[MAX<<2]={0};
long long int num[MAX];void Build(long long int t,int l,int r)             //以1为根节点建立线段树[l,r]
{Tree[t].left=l,Tree[t].right=r;if(Tree[t].left==Tree[t].right){Tree[t].sum=num[Tree[t].right];Tree[t].add=0;return ;}int mid=MID(Tree[t].left,Tree[t].right);Build(L(t),l,mid);Build(R(t),mid+1,r);Tree[t].sum=Tree[L(t)].sum+Tree[R(t)].sum;
}void Insert(long long int t,int l,int r,long long int m)   //向区间[l,r]插入m
{if(Tree[t].left==l&&Tree[t].right==r){Tree[t].sum+=(Tree[t].right-Tree[t].left+1)*m;Tree[t].add+=m;return ;}if(Tree[t].add!=0)                   //无论是插入还是查询都要更新lazy{Tree[L(t)].sum+=(Tree[L(t)].right-Tree[L(t)].left+1)*Tree[t].add;Tree[R(t)].sum+=(Tree[R(t)].right-Tree[R(t)].left+1)*Tree[t].add;Tree[L(t)].add+=Tree[t].add;Tree[R(t)].add+=Tree[t].add;Tree[t].add=0;}int mid=MID(Tree[t].left,Tree[t].right);if(l>mid){Insert(R(t),l,r,m);}else if(r<=mid){Insert(L(t),l,r,m);}else{Insert(L(t),l,mid,m);Insert(R(t),mid+1,r,m);}Tree[t].sum=Tree[L(t)].sum+Tree[R(t)].sum;
}long long int Query(long long int t,int l,int r)     //查询区间[a,b]
{if(Tree[t].left==l&&Tree[t].right==r){return Tree[t].sum;}if(Tree[t].add!=0)                 //lazy更新{Tree[L(t)].sum+=(Tree[L(t)].right-Tree[L(t)].left+1)*Tree[t].add;Tree[R(t)].sum+=(Tree[R(t)].right-Tree[R(t)].left+1)*Tree[t].add;Tree[L(t)].add+=Tree[t].add;Tree[R(t)].add+=Tree[t].add;Tree[t].add=0;}int mid=MID(Tree[t].left,Tree[t].right);if(l>mid){return Query(R(t),l,r);}else if(r<=mid){return Query(L(t),l,r);}else{return Query(L(t),l,mid)+Query(R(t),mid+1,r);}Tree[t].sum=Tree[L(t)].sum+Tree[R(t)].sum;   //更新结点: 结点的值=左子树+右子树
}int main()
{char ch;int n,a,b,c;long long int i,m;scanf("%d%lld",&n,&m);for(i=1;i<=n;i++)         //初始化输入scanf("%lld",&num[i]);Build(1,1,n);for(i=1;i<=m;i++){getchar();scanf("%c",&ch);if(ch=='C'){scanf("%d%d%d",&a,&b,&c);Insert(1,a,b,c);            //区间[a,b]都加上c}else{scanf("%d%d",&a,&b);printf("%lld\n",Query(1,a,b));  //查询区间[a,b]的和}}return 0;
}

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