题解

$\color{red}{题解传送门}$

题目

题目传送门

题解

• $\gcd(a,b)=\gcd(a, b-a)$
• $\gcd(a,b,c) = \gcd(a,b-a, c-b)$
• $\gcd(a_1,a_2,\cdots,a_n) = \gcd(a_1, a_2-a_1, a_3-a_2,\cdots,a_n-a_{n-1})$
• 知道了这个性质我们可以开一个$B[\ \ ]$数组，表示原序列的差分序列，用线段数维护$B$的区间最大公约数
• 对于询问“Q l r”等于求出$\gcd(a[l], query(1,l+r, r))$
• 对于询问”C l r d”等于$B[l]$加上$d$，$B[r+1]$减去$d$，只需两次线段数的单点修改即可(线段树单点修改模板)，对于原序列中$A$的值，我们可以用一个“区间修改+单点查询”的树状数组来维护(树状数组区间修改+单点查询模板)

code

#include <bits/stdc++.h> 
using namespace std; 
const int maxn = 5e5 + 100; 
typedef long long LL; 

template <typename T> 
inline void read(T &s) {
    s = 0; 
    T w = 1, ch = getchar(); 
    while (!isdigit(ch)) { if (ch == '-') w = -1; ch = getchar(); }
    while (isdigit(ch)) { s = (s << 1) + (s << 3) + (ch ^ 48); ch = getchar(); }
    s *= w; 
}

LL n, m; 
LL a[maxn], b[maxn], c[maxn]; // 原数组，差分数组，增量数组（树状数组） 
struct node {
    LL l, r; 
    LL dat; 
} t[maxn * 4]; 

inline LL gcd(LL x, LL y) { return y ? gcd(y, x % y) : x; }

inline void build(LL p, LL l, LL r) {
    t[p].l = l, t[p].r = r; 
    if (l == r) { t[p].dat = b[l]; return ; }
    LL mid = (l + r)>>1; 
    build(p<<1, l, mid); 
    build(p<<1|1, mid + 1, r); 
    t[p].dat = gcd(t[p<<1].dat, t[p<<1|1].dat); 
}

inline void change(LL p, LL x, LL val) {
    if (t[p].l == t[p].r) { t[p].dat += val; return ; }
    LL mid = (t[p].l + t[p].r)>>1; 
    if (x <= mid) change(p<<1, x, val); 
    else change(p<<1|1, x, val); 
    t[p].dat = gcd(t[p<<1].dat, t[p<<1|1].dat); 
}

inline LL query(LL p, LL l, LL r) {
    if (l <= t[p].l && r >= t[p].r) return t[p].dat; 
    LL mid = (t[p].l + t[p].r)>>1; 
    LL val = 0; 
    if (l <= mid) val = gcd(val, query(p<<1, l, r));  
    if (r > mid) val = gcd(val, query(p<<1|1, l, r)); 
    return abs(val); 
}

inline LL lowbit(LL x) { return x & -x; }

inline void add(LL x, LL val) {
    for (; x <= n; x += lowbit(x))
        c[x] += val; 
}

inline LL ask(LL x) {
    LL ans = 0; 
    for (; x; x -= lowbit(x))
        ans += c[x]; 
    return ans; 
}

int main() {
    read(n), read(m); 
    for (LL i = 1; i <= n; ++i) { 
        read(a[i]); 
        b[i] = a[i] - a[i-1]; 
    }
    build(1, 1, n); 
    for (LL i = 1; i <= m; ++i) {
        char opt[2]; LL l, r, val; 
        scanf("%s", opt); 
        if (opt[0] == 'Q') {
            read(l), read(r); 
            LL now = a[l] + ask(l); 
            printf("%lld\n", gcd(now, query(1, l+1, r))); 
        } else {
            read(l), read(r), read(val); 
            add(l, val); 
            add(r + 1, -val); 
            change(1, l, val); 
            if (r < n) change(1, r + 1, -val); 
        }
    }
    return 0; 
}