[学习笔记] 树状数组区间加+区间求和

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记bi=ai−ai−1,ci=(i−1)×bi，则：
$\sum i = 1 n a i$ $= a 1 + a 2 + \dots + a n$ $= \sum i = 1 1 b i + \sum i = 1 2 b i + \dots + \sum i = 1 n b i$ $= n \times \sum i = 1 n b i - \sum i = 1 n (i - 1) \times b i$ $= n \times \sum i = 1 n b i - \sum i = 1 n c i$
我们用两个树状数组分别维护bi,ci
1.令区间[l,r]加上x，即为 $b l + = x, b r + 1 - = x$ $c l + = (l - 1) \times x, c r + 1 - = r \times x$
2.询问区间[l,r]的和，即为 $r \times \sum i = 1 r b i - \sum i = 1 r c i - (l - 1) \times \sum i = 1 l - 1 b i + \sum i = 1 l - 1 c i$
代码

const int N = 1e5 + 5;ll b[N], c[N]; int a[N], n, m;inline void Modify(ll *s, int x, ll y){    for (; x <= n; x += x & -x)        s[x] += y;}inline ll Query(ll *s, int x){    ll res = 0;    for (; x; x ^= x & -x)        res += s[x];    return res;}int main(){    n = get(); m = get(); int k, l, r, x;    for (int i = 1; i <= n; ++i)    {        a[i] = get();        Modify(b, i, a[i] - a[i - 1]);        Modify(c, i, 1ll * (i - 1) * (a[i] - a[i - 1]));    }    while (m--)    {        k = get(); l = get(); r = get();        if (k & 1)        {            x = get();            Modify(b, l, x);             Modify(b, r + 1, -x);            Modify(c, l, 1ll * (l - 1) * x);            Modify(c, r + 1, -1ll * r * x);         }        else         {            --l;            put(Query(b, r) * r - Query(c, r)              - Query(b, l) * l + Query(c, l));            putchar('\n');         }    }}

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