poj 3468A Simple Problem with Integers

Description

You have N integers, A₁, A₂, ... , A_N. 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, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a+1, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a, A_a+1, ... , A_b.

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

The sums may exceed the range of 32-bit integers.

这题是线段树成段更新，看别人代码想了很久。

思路：开一个结构体，含有左右边界l,r，这段区间线段总和sum，更新标志ans（整段区间内每个数要加的数值）。

每次一个区间更新（即加一个数value)的时候，从第一个线线段开始向下判断，如果更新的区间刚好是这条线段的区间，那么直接加上更新的数值value并返回，否则整段区间的sum值变为sum+(b-a+1)*value，再在子节点中找直到找到区间大小加好符合的时候，ans=ans+value,返回。

每一次询问，从第一个线段开始向下，如果区间刚好符合，那么返回区间的sum否则把这条线段的更新标志往子节点传，同时这条线段的更新标志变为0.这里我采用的是每一次更新就把每根点段的总和都保存在sum中，这样询问的时候就不用麻烦的加上b[i].ans*(b[i].r-b[i].l+1).

#include<stdio.h>
#include<string.h>
#define maxn 100005
#define ll long long
char str[10];
ll a[maxn];
struct node
{
	ll l,r,sum,ans;
}b[4*maxn];

void build(ll l,ll r,ll i)
{
    ll mid;
    b[i].l=l;
    b[i].r=r;
    b[i].ans=0;
    if(b[i].l==b[i].r)
    {
       b[i].sum=a[l];
       return;
    }
    mid=(l+r)/2;
    build(l,mid,2*i);
    build(mid+1,r,i*2+1);
    b[i].sum=b[i*2].sum+b[i*2+1].sum;
}

void pushdown(int i)
{
    if(b[i].ans){
        b[i*2].ans+=b[i].ans;
        b[i*2+1].ans+=b[i].ans;
        b[i*2].sum+=b[i].ans*(b[i*2].r-b[i*2].l+1);
        b[i*2+1].sum+=b[i].ans*(b[i*2+1].r-b[i*2+1].l+1);
        b[i].ans=0;
    }
}


void add(ll l,ll r,ll value,ll i)
{
	ll mid;
	if(b[i].l==l && b[i].r==r)
	{
	   b[i].ans=b[i].ans+value;
	   b[i].sum+=(b[i].r-b[i].l+1)*value;
	   return;
	}
	pushdown(i);
	b[i].sum+=(r-l+1)*value; //这一句写了，下面第二句就不用写了，是同一个意思
	mid=(b[i].l+b[i].r)/2;
	if(l>mid)
	add(l,r,value,i*2+1);
	else if(r<=mid)
	add(l,r,value,i*2);
	else
	{
	   add(l,mid,value,i*2);
	   add(mid+1,r,value,i*2+1);
	}
	//b[i].sum=b[i*2].sum+b[i*2+1].sum;  ---2
}

ll question(ll l,ll r,ll i)
{
	ll mid;
	if(b[i].l==l && b[i].r==r)
	{
	   return b[i].sum;
	}
	pushdown(i);
	mid=(b[i].l+b[i].r)/2;
	if(l>mid)
	return question(l,r,i*2+1);
	else if(r<=mid)
	return question(l,r,i*2);
	else if(l<=mid && r>mid)
	return question(l,mid,i*2)+question(mid+1,r,i*2+1);

}

int main()
{
	ll n,m,c,d,e,i,j;
	while(scanf("%lld%lld",&n,&m)!=EOF)
	{
		for(i=1;i<=n;i++)
		scanf("%lld",&a[i]);
		build(1,n,1);
		while(m--)
		{
			scanf("%s",str);
			if(str[0]=='Q')
			{
				scanf("%lld%lld",&c,&d);
				printf("%lld
",question(c,d,1));
			}
			else if(str[0]=='C')
			{
				scanf("%lld%lld%lld",&c,&d,&e);
				add(c,d,e,1);
			}
		}
	}
	return 0;
}