poj 3468

经典线段树，大彻大悟了！！！

代码：

#include<iostream>
#include<fstream>

using namespace std;

struct e{
	int left,right;
	long long add;
	long long sum;
};

e tree[400011];
long long a[100001];

long long build(int s,int t,int num){
	long long i,j;
	tree[num].left=s;
	tree[num].right=t;
	tree[num].add=0;
	if(s!=t)
	{
		i=build(s,(s+t)/2,num*2);
		j=build((s+t)/2+1,t,num*2+1);
	}
	else
	{
		tree[num].sum=a[s];
		return(a[s]);
	}
	tree[num].sum=i+j;
	return(tree[num].sum);
}

void insert(int s,int t,int add,int num){
	
	if(s<=tree[num].left&&tree[num].right<=t)
	{
		tree[num].add+=add;
		tree[num].sum+=add*(tree[num].right-tree[num].left+1);
		return;
	}
	if(tree[num].add){
		tree[num*2].sum+=tree[num].add*(tree[num*2].right-tree[num*2].left+1);
		tree[num*2+1].sum+=tree[num].add*(tree[num*2+1].right-tree[num*2+1].left+1);
		tree[num*2].add+=tree[num].add;
		tree[num*2+1].add+=tree[num].add;
		tree[num].add=0;
	}
	if(s<=(tree[num].left+tree[num].right)/2)
		insert(s,t,add,num*2);
	if(t>(tree[num].left+tree[num].right)/2)
		insert(s,t,add,num*2+1);
	tree[num].sum=tree[num*2].sum+tree[num*2+1].sum;
	
}

long long search(int s,int t,int num){
	long long i=0,j=0;
	if(s<=tree[num].left&&tree[num].right<=t)
	{
		return(tree[num].sum);
	}
	if(tree[num].add){
		tree[num*2].sum+=tree[num].add*(tree[num*2].right-tree[num*2].left+1);
		tree[num*2+1].sum+=tree[num].add*(tree[num*2+1].right-tree[num*2+1].left+1);
		tree[num*2].add+=tree[num].add;
		tree[num*2+1].add+=tree[num].add;
		tree[num].add=0;
	}
	if(s<=(tree[num].left+tree[num].right)/2)
	{
		i=search(s,t,num*2);
		
	}
	if(t>(tree[num].left+tree[num].right)/2)
	{
		j=search(s,t,num*2+1);
		
	}
	
	return(i+j);
}


int main(){
	//ifstream cin("in.txt");
	int i,j,k,n,m,p,t;
	char s;
	scanf("%d%d",&n,&m);
	for(i=1;i<=n;i++)
		scanf("%lld",&a[i]);
	build(1,n,1);
	for(i=1;i<=m;i++)
	{
		scanf("\n%c",&s);
		if(s=='C')
		{
			scanf("%d%d%d",&j,&k,&p);
			if(j>k) {t=j;j=k;k=t;}
			insert(j,k,p,1);
		}
		else
		{
			scanf("%d%d",&j,&k);
			if(j>k) {t=j;j=k;k=t;}
			cout<<search(j,k,1)<<endl;
		}
	}

	return(0);
}