POJ--3468 A Simple Problem with Integers(树状数组维护)

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.

　　　　题意：Q：求L到R的和。C：L,R的数都加x

　　　　解析：关于差分：https://www.cnblogs.com/liyexin/p/11014218.html。下面c1[]和c2[]都是采用了差分的思想。

　　　　　　　设原数组的前缀和：sum[]。数组大小为n。

　　　　　　　设c1[i]表示i~n，每个数字+x。要[L,R]加x，那么有c1[L]+=x，c2[R+1]-=x。

　　　　　　　对于sum[x]，c1[i]对其贡献为(x-i+1)。那么求1-x的和，有：sum[1]+sum[2]+....+c1[1]*x+c1[2]*(x-1)+.....c1[x]*1。用se表示累加求和1~x，那么原式就是se(sum[x])+se(c1[i]*(x-i+1))==sum[x]+(x+1)*se(c1[i])-se(c1[i]*i)。那么se的意思也是前缀和了，所以用c1[i]为i的前缀和，c2[i]为c2[i]*i的前缀和。那么更新时，需要更新两个数组：

            update(a,c,c1);
            update(b+1,-c,c1);
            update(a,a*c,c2);
            update(b+1,-(b+1)*c,c2);

　　　　　　　　求和时，套一下上面的公式：

            ll a,b;
            scanf("%lld%lld",&a,&b);
            ll x1=sum[b]+(b+1)*get(b,c1)-get(b,c2);
            ll x2=sum[a-1]+a*get(a-1,c1)-get(a-1,c2);
            cout<<x1-x2<<endl;

#include<iostream>
#include<cstdio>
#include<cstring>
#include<map>
using namespace std;
typedef long long ll;
const int maxn = 1e5+10;
ll sum[maxn],c1[maxn],c2[maxn];
ll n,q;
ll lowbit(ll x)
{
    return x&(-x);
}
void update(ll id,ll x,ll *c)
{
    for(int i=id;i<=n;i+=lowbit(i))
    {
        c[i]+=x;
    }
}
ll get(ll id,ll *c)
{
    ll sum=0;
    for(int i=id;i>0;i=i-lowbit(i))
    {
        sum+=c[i];
    }
    return sum;
}
int main()
{
    scanf("%lld%lld",&n,&q);
    ll x,ans=0;
    for(int i=1;i<=n;i++)
    {
        scanf("%lld",&x);
        ans+=x;
        sum[i]=ans;
    }
    while(q--)
    {
        char s[20];
        scanf("%s",s);
        if(s[0]=='C')
        {
            ll a,b,c;
            scanf("%lld%lld%lld",&a,&b,&c);
            update(a,c,c1);
            update(b+1,-c,c1);
            update(a,a*c,c2);
            update(b+1,-(b+1)*c,c2);
        }
        else
        {
            ll a,b;
            scanf("%lld%lld",&a,&b);
            ll x1=sum[b]+(b+1)*get(b,c1)-get(b,c2);
            ll x2=sum[a-1]+a*get(a-1,c1)-get(a-1,c2);
            cout<<x1-x2<<endl;
        }
    }
}