POJ3468 A Simple Problem with Integers 线段树成段更新

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A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 89719		Accepted: 27930
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

Hint

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

#include <iostream>
#include <stdio.h>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#define ll __int64
using namespace std;

ll a[4*100009];
ll lazy[4*100009];

void pushdown(int le,int mid,int ri,int num)
{
    if(lazy[num])
    {
        a[num*2]+=lazy[num]*(mid-le+1);
        a[num*2+1]+=lazy[num]*(ri-mid);   //注意计算个数
        lazy[num*2]+=lazy[num];
        lazy[num*2+1]+=lazy[num];
        lazy[num]=0;
    }
}

void buildtree(int le,int ri,int num)
{
    if(le==ri)
    {
        scanf("%I64d",&a[num]);
        return;
    }
    int mid=(le+ri)/2;
    buildtree(le,mid,num*2);
    buildtree(mid+1,ri,num*2+1);
    a[num]=a[num*2]+a[num*2+1];
}

void add(int le,int ri,int num,int x,int y,ll z)
{
    if(x<=le&&y>=ri)
    {
        a[num]+=(ri-le+1)*z;
        lazy[num]+=z;
        return;
    }
    int mid=(le+ri)/2;
    pushdown(le,mid,ri,num);

    if(x<=mid)
        add(le,mid,num*2,x,y,z);
    if(y>mid)
        add(mid+1,ri,num*2+1,x,y,z);

    a[num]=a[num*2]+a[num*2+1];
}

ll query(int le,int ri,int num,int x,int y)
{
    if(x<=le && y>=ri)
    {
        return a[num];
    }
    int mid=(le+ri)/2;
    pushdown(le,mid,ri,num);
    ll ans=0;
    if(x<=mid)
        ans+=query(le,mid,num*2,x,y);
    if(y>mid)
        ans+=query(mid+1,ri,num*2+1,x,y);

    return ans;
}

int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        memset(a,0,sizeof a);
        memset(lazy,0,sizeof lazy);
        buildtree(1,n,1);

        int x,y;
        ll z;
        while(m--)
        {
            char s[10];
            scanf("%s",s);
            if(s[0]=='Q')
            {
                scanf("%d%d",&x,&y);
                printf("%I64d\n",query(1,n,1,x,y));
            }
            else
            {
                 scanf("%d%d%I64d",&x,&y,&z);
                 add(1,n,1,x,y,z);
            }
        }
    }
    return 0;
}