Codevs1082 線段樹練習三(樹狀數組)

終於抽出時間學了一發樹狀數組的區間修改

用c[i]=a[i]−a[i−1]
很容易得到:a[i]=∑i=1ci
s[i]=(i−1)⋅c[i]
那麼到n的答案爲
ans=(n−1)⋅∑i=1ci−∑i=1si

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fod(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
typedef long long ll;
const int N=2e5+10;
int n,m;
ll delta[N],deltai[N],sum[N],a[N];
#define lowbit(x) ((x)&(-x))
ll query(ll *array,int pos)  
{  
    ll temp = 0ll;  
    while(pos > 0)  {  
        temp += array[pos];  
        pos -= lowbit(pos);  
    }  
    return temp;  
}  
void update(ll *array,int pos,int x)  
{  
    while(pos <= N)  
    {  
        array[pos] += x;  
        pos += lowbit(pos);  
    }  
}  
int main()
{
    scanf("%d",&n);
    fo(i,1,n) {scanf("%lld",&a[i]);sum[i]=sum[i-1]+a[i];}
    scanf("%d",&m);
    for(int op,l,r,x,i=1;i<=m;i++) {
        scanf("%d",&op);
        if(op==1) {
            scanf("%d%d%d",&l,&r,&x);
            update(delta,l,x);
            update(delta,r+1,-x);
            update(deltai,l,x*l);
            update(deltai,r+1,-x*(r+1));
        }
        else {
            scanf("%d%d",&l,&r);
            ll suml=sum[l-1]+l*query(delta,l-1)-query(deltai,l-1);
            ll sumr=sum[r]+(r+1)*query(delta,r)-query(deltai,r);
            printf("%lld\n",sumr-suml);
        }
    }
    return 0;
}