BZOJ3212: Pku3468 A Simple Problem with Integers

3212: Pku3468 A Simple Problem with Integers

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 810  Solved: 354
[Submit][Status]

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. 

Source

题解：

想了想树状数组如何维护区间修改，区间求和，觉得貌似很简单？

差分了之后 用树状数组维护 a[i] 以及  i*a[i] 即可

sum(1,n)=(n+1)*sigma(a[i])-sigma(i*a[i])

靠着树状数组刷进了第一页。。。

代码：

#include<cstdio>

#include<cstdlib>

#include<cmath>

#include<cstring>

#include<algorithm>

#include<iostream>

#include<vector>

#include<map>

#include<set>

#include<queue>

#include<string>

#define inf 1000000000

#define maxn 100000+1000

#define maxm 500+100

#define eps 1e-10

#define ll long long

#define pa pair<int,int>

#define for0(i,n) for(int i=0;i<=(n);i++)

#define for1(i,n) for(int i=1;i<=(n);i++)

#define for2(i,x,y) for(int i=(x);i<=(y);i++)

#define for3(i,x,y) for(int i=(x);i>=(y);i--)

#define mod 1000000007

using namespace std;

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=10*x+ch-'0';ch=getchar();}

    return x*f;

}
ll s[2][maxn],n,m;
inline void add(int k,int x,ll y)
{
    for(;x<=n;x+=x&(-x))s[k][x]+=y;
}
inline ll sum(int k,int x)
{
    ll t=0;
    for(;x;x-=x&(-x))t+=s[k][x];
    return t;
}

int main()

{

    freopen("input.txt","r",stdin);

    freopen("output.txt","w",stdout);

    n=read();m=read();
    ll x=0,y,z;
    for(ll i=1;i<=n;i++)
    {
        y=read();
        add(0,i,y-x);
        add(1,i,i*(y-x));
        x=y;
    }
    while(m--)
    {
        char ch[2];
        scanf("%s",ch);
        if(ch[0]=='C')
        {
            x=read();y=read();z=read();
            add(0,x,z);add(1,x,x*z);
            add(0,y+1,-z);add(1,y+1,(-z)*(y+1));
        }
        else
        {
            x=read();y=read();
            ll sum1=(x)*sum(0,x-1)-sum(1,x-1);
            ll sum2=(y+1)*sum(0,y)-sum(1,y);
            printf("%lld
",sum2-sum1);
        }
    }

    return 0;

}