poj 3468 A Simple Problem with Integers

A Simple Problem with Integers
Time Limit: 5000MS Memory Limit: 131072K
Total Submissions: 100482 Accepted: 31341
Case Time Limit: 2000MS

Description

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

Sample Output

4
55
9
15

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

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【分析】
继续刷线段树….
区间增减+区间求和
现在是要用到懒标记这个东西了

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【代码】

//poj 3468 A Simple Problem with Integers 
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
#define M(a) memset(a,0,sizeof a)
#define fo(i,j,k) for(i=j;i<=k;i++) 
using namespace std;
const int mxn=100005;
struct node {int l,r;ll sum,lazy;} t[4*mxn];
ll n,m,T,L,R,k;
ll a[mxn];
char s[15];
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-') f=-1;ch=getchar();} 
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x*f;
}
inline void build(int num,int l,int r)
{
    t[num].l=l,t[num].r=r;
    if(l==r)
    {
        t[num].sum=a[l];
        return;
    }
    int mid=(l+r)>>1;
    build(num<<1,l,mid);
    build((num<<1)+1,mid+1,r);
    t[num].sum=t[num<<1].sum+t[(num<<1)+1].sum;
}
inline void pushdown(int num)
{
    if(t[num].lazy==0) return;
    t[num<<1].lazy+=t[num].lazy;
    t[num<<1].sum+=(t[num<<1].r-t[num<<1].l+1)*t[num].lazy;
    t[(num<<1)+1].lazy+=t[num].lazy;
    t[(num<<1)+1].sum+=(t[(num<<1)+1].r-t[(num<<1)+1].l+1)*t[num].lazy;
    t[num].lazy=0;
}
inline void add(int num)
{
    if(L<=t[num].l && t[num].r<=R)
    {
        t[num].lazy+=k;
        t[num].sum+=(t[num].r-t[num].l+1)*k;
        return;
    }
    pushdown(num);
    int mid=(t[num].l+t[num].r)>>1;
    if(L<=mid) add(num<<1);
    if(mid<R) add((num<<1)+1);
    t[num].sum=(t[num<<1].sum+t[(num<<1)+1].sum);
}
inline ll query(int num)
{
    ll ans=0;
    if(L<=t[num].l && t[num].r<=R)
      return t[num].sum;
    pushdown(num);
    int mid=(t[num].l+t[num].r)>>1;
    if(L<=mid) ans+=query(num<<1);
    if(mid<R) ans+=query((num<<1)+1);
    return ans;
}
int main()
{
    int i,j;
    n=read(),m=read();
    fo(i,1,n) a[i]=read();
    build(1,1,n);
    while(m--)
    {
        scanf("%s",s);
        if(s[0]=='C')
        {
            L=read(),R=read(),k=read();
            add(1);
        }
        else
        {
            L=read(),R=read();
            printf("%lld\n",query(1));
        }
    }
    return 0;
}