树状数组的区间修改、区间查询

设原数组第$i$位为$a[i]$，$a[0]=0$，$d[i]=a[i]-a[i-1]$

那么$a[x]={\sum }_{i=1}^{x}d[i]$

${\sum }_{i=1}^{x}a[i]={\sum }_{i=1}^x{\sum }_{j=1}^{i}d[j]={\sum }_{i=1}^x(x-i+1)d[i]=(x+1){\sum }_{i=1}^{x}d[i]-{\sum }_{i=1}^{x}i\ast d[i]$

即原数组差分后，维护${\sum }_{i=1}^{x}d[i]$和${\sum }_{i=1}^{x}i\ast d[i]$即可

inline int lowbit (int x) {return x & (-x);}
inline void update (int pos, ll val) {
    for (int i = pos; i <= n; i += lowbit (i)) {
        c1[i] += val; c2[i] += 1ll * pos * val ;
    }
}
inline ll cal (int pos) {
    ll res = 0 ;
    for (int i = pos; i; i -= lowbit (i))
        res += c1[i] * (pos + 1) - c2[i] ;
    return res ;
}