数列分块入门 7

拼搏现实的明天。 2022-01-07 05:19 150阅读 0赞

# 分块训练 #

两个tag，优先算乘法的tag。

暴力更新的时候要把乘法标记和加法标记都清空。

#include <bits/stdc++.h>
    #define INF 0x3f3f3f3f
    #define full(a, b) memset(a, b, sizeof a)
    using namespace std;
    typedef long long ll;
    inline int lowbit(int x){ return x & (-x); }
    inline int read(){
        int X = 0, w = 0; char ch = 0;
        while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
        while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
        return w ? -X : X;
    }
    inline int gcd(int a, int b){ return b ? gcd(b, a % b) : a; }
    inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
    template<typename T>
    inline T max(T x, T y, T z){ return max(max(x, y), z); }
    template<typename T>
    inline T min(T x, T y, T z){ return min(min(x, y), z); }
    template<typename A, typename B, typename C>
    inline A fpow(A x, B p, C lyd){
        A ans = 1;
        for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
        return ans;
    }
    const int N = 100005;
    const int MOD = 10007;
    ll a[N], add[N], mul[N];
    int n, t, lt[N], rt[N], pos[N];
    
    void adder(int l, int r, ll d){
        d %= MOD;
        int p = pos[l], q = pos[r];
        if(p == q){
            for(int i = lt[p]; i <= rt[p]; i ++)
                a[i] = ((a[i] * mul[p]) % MOD + add[p]) % MOD;
            for(int i = l; i <= r; i ++)
                a[i] = (a[i] + d) % MOD;
            add[p] = 0, mul[p] = 1;
        }
        else{
            for(int i = p + 1; i <= q - 1; i ++)
                add[i] = (add[i] + d) % MOD;
            for(int i = lt[p]; i <= rt[p]; i ++)
                a[i] = ((a[i] * mul[p]) % MOD + add[p]) % MOD;
            for(int i = l; i <= rt[p]; i ++)
                a[i] = (a[i] + d) % MOD;
            for(int i = lt[q]; i <= rt[q]; i ++)
                a[i] = ((a[i] * mul[q]) % MOD + add[q]) % MOD;
            for(int i = lt[q]; i <= r; i ++)
                a[i] = (a[i] + d) % MOD;
            add[p] = add[q] = 0, mul[p] = mul[q] = 1;
        }
    }
    
    void muler(int l, int r, ll d){
        d %= MOD;
        int p = pos[l], q = pos[r];
        if(p == q){
            for(int i = lt[p]; i <= rt[p]; i ++)
                a[i] = ((a[i] * mul[p]) % MOD + add[p]) % MOD;
            for(int i = l; i <= r; i ++)
                a[i] = (a[i] * d) % MOD;
            add[p] = 0, mul[p] = 1;
        }
        else{
            for(int i = p + 1; i <= q - 1; i ++)
                mul[i] = (mul[i] * d) % MOD, add[i] = (add[i] * d) % MOD;
            for(int i = lt[p]; i <= rt[p]; i ++)
                a[i] = ((a[i] * mul[p]) % MOD + add[p]) % MOD;
            for(int i = l; i <= rt[p]; i ++)
                a[i] = (a[i] * d) % MOD;
            for(int i = lt[q]; i <= rt[q]; i ++)
                a[i] = ((a[i] * mul[q]) % MOD + add[q]) % MOD;
            for(int i = lt[q]; i <= r; i ++)
                a[i] = (a[i] * d) % MOD;
            add[p] = add[q] = 0, mul[p] = mul[q] = 1;
        }
    }
    
    ll query(int k){
        return (a[k] * mul[pos[k]] % MOD + add[pos[k]] % MOD) % MOD;
    }
    
    int main(){
    
        //freopen("data.txt", "r", stdin);
    
        ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
    
        cin >> n;
        for(int i = 1; i <= n; i ++) cin >> a[i];
        t = (int)sqrt(n);
        for(int i = 1; i <= t; i ++){
            lt[i] = (i - 1) * t + 1;
            rt[i] = i * t;
        }
        if(rt[t] < n) t ++, lt[t] = rt[t - 1] + 1, rt[t] = n;
        for(int i = 1; i <= t; i ++){
            mul[i] = 1, add[i] = 0;
            for(int j = lt[i]; j <= rt[i]; j ++){
                pos[j] = i;
            }
        }
        for(int i = 1; i <= n; i ++){
            int opt, l, r; ll c;
            cin >> opt >> l >> r >> c;
            if(opt == 0) adder(l, r, c);
            else if(opt == 1) muler(l, r, c);
            else cout << query(r) << endl;
        }
        return 0;
    }

转载于:https://www.cnblogs.com/onionQAQ/p/10878226.html