A Simple Problem with Integers

Posted 2020-08-16 勿忘初心0924

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 99895		Accepted: 31162
Case Time Limit: 2000MS

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.

Source

POJ Monthly--2007.11.25, Yang Yi

/*
手敲模板结果各种手残！
*/
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#define N 100010
#define ll long long
#define lson i*2,l,m
#define rson i*2+1,m+1,r
using namespace std;
/*
线段树区间更新的时候将更新值存入数组add数组中，等你需要的再去加到sum数组中
*/
ll sum[N*4];
ll add[N*4];
void pushdown(int i,int num)///向下更新
{
    if(add[i])///等于0的话没有更新的必要了
    {
        sum[i*2] += add[i]*(num-(num/2));
        sum[i*2+1] += add[i]*(num/2);
        add[i*2]+=add[i];
        add[i*2+1]+=add[i];
        add[i]=0;///这个节点的信息更新完了，那么相应的存在add数组中的东西就没有了
    }
}
void pushup(int i)///向下更新
{
    sum[i]=sum[i*2]+sum[i*2+1];
}
void build(int i,int l,int r)
{
    add[i]=0;///将每个节点更新的值初始化为0
    if(l==r)
    {
        scanf("%lld",&sum[i]);
        //cout<<sum[i]<<" ";
        return ;
    }
    int m=(l+r)/2;
    build(lson);
    build(rson);
    pushup(i);
}
void update(int ql,int qr,int val,int i,int l,int r)
{
    if(ql<=l&&r<=qr)
    {
        add[i]+=val;
        sum[i]+=(ll)val*(r-l+1);
        return ;
    }
    pushdown(i,r-l+1);
    int m=(l+r)/2;
    if(ql<=m) update(ql,qr,val,lson);
    if(m<qr) update(ql,qr,val,rson);
    pushup(i);
}
ll query(int ql,int qr,int i,int l,int r)
{
    //cout<<"l="<<l<<" r="<<r<<endl;
    if(ql<=l&&r<=qr)
    {
        return sum[i];
    }
    pushdown(i,r-l+1);
    int m=(l+r)/2;
    ll cur=0;
    if(ql<=m) cur+=query(ql,qr,lson);
    if(m<qr) cur+=query(ql,qr,rson);
    return cur;
}
int main()
{
    //freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin);
    int n,q;
    char op;
    while(scanf("%d%d",&n,&q)!=EOF)
    {
        //cout<<n<<" "<<m<<endl;
        build(1,1,n);
        //for(int i=1;i<=(n*(n+1)/2);i++)
        //    cout<<sum[i]<<" ";
        //cout<<endl;
        //cout<<endl;
        int a,b,c;
        getchar();
        while(q--)
        {
            scanf("%c",&op);
            //cout<<op<<" ";
            if(op==‘Q‘)
            {
                scanf("%d%d",&a,&b);
                //cout<<a<<" "<<b<<endl;
                //cout<<"Q"<<endl;
                printf("%lld\n",query(a,b,1,1,n));
            }
            else
            {
                scanf("%d%d%d",&a,&b,&c);
                //cout<<"C"<<endl;
                update(a,b,c,1,1,n);
            }
            getchar();
        }
    }
    return 0;
}