A Simple Problem with Integers

来源poj3468

You have N integers, A1, A2, ... , AN. 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, ... , 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.

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

区间修改的线段树

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include <iomanip>
#include<cmath>
#include<float.h> 
#include<string.h>
#include<algorithm>
#include<vector>
#include<queue>
#define sf scanf
#define pf printf
#define scf(x) scanf("%d",&x)
#define scff(x,y) scanf("%d%d",&x,&y)
#define scfff(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define prf(x) printf("%d\n",x) 
#define mm(x,b) memset((x),(b),sizeof(x))
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=a;i>=n;i--)
typedef long long ll;
const ll mod=1e9+7;
const double eps=1e-8;
const int inf=0x3f3f3f3f;
using namespace std;
const double pi=acos(-1.0);
const int N=1e6+10;
struct TREE
{
	ll l,r;
	ll sum;
}tr[N];
ll a[N];
ll add[N];
void built_tree(ll x,ll y,ll i)
{
	tr[i].l =x;tr[i].r =y;
	if(x==y)
	tr[i].sum =a[x];
	else
	{
		ll mid=(x+y)>>1;
		built_tree(x,mid,i<<1);
		built_tree(mid+1,y,i<<1|1);
		tr[i].sum=tr[i<<1].sum+tr[i<<1|1].sum; 
	}
}
void pushdown(int i)//下放惰性标记 
{
	ll lc=i<<1,rc=i<<1|1;
	tr[lc].sum +=(tr[lc].r -tr[lc].l +1)*add[i];
	tr[rc].sum +=(tr[rc].r -tr[rc].l +1 )*add[i];
	add[lc]+=add[i];
	add[rc]+=add[i];
	add[i]=0;
}
void update_tree(ll x,ll y,ll k,int i)
{
	ll lc=i<<1;
	ll rc=i<<1|1;
	if(tr[i].l>y||tr[i].r<x) return; //如果不属于，则返回
	if(x <= tr[i].l &&tr[i].r <=y)
	{
		tr[i].sum+=(tr[i].r -tr[i].l +1)*k;//加至此处不继续往下加 
		add[i]+=k;//存惰性标记 
	}else
	{
		if(add[i])
			pushdown(i);
		update_tree(x,y,k,lc);
		update_tree(x,y,k,rc);
		tr[i].sum =tr[lc].sum+tr[rc].sum; 
	} //查找 
}

ll query(ll x,ll y,int i)
{
	ll lc=i<<1,rc=i<<1|1;
	if(x <= tr[i].l && tr[i].r <= y) return tr[i].sum;
	if(x > tr[i].r|| y < tr[i].l) return 0;
	if(add[i])
		pushdown(i);
	return query(x,y,lc)+query(x,y,rc);
}

int main()
{
	int n,q;
	scff(n,q);
	rep(i,1,n+1)
	sf("%lld",&a[i]);
	built_tree(1,n,1);
	while(q--)
	{
		char c;
		cin>>c;ll x,y,z;
		if(c=='Q')
		{
			sf("%lld%lld",&x,&y);
			pf("%lld\n",query(x,y,1));
		}else
		{
			sf("%lld%lld%lld",&x,&y,&z);
			update_tree(x,y,z,1);
		}
	}
	return 0;
}