• 数列分块入门 4


    分块训练

    分好块维护一个sum数组,中间区间只更新加标记,两边暴力更新即可。

    #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 = 50005;
    ll a[N], add[N], sum[N];
    int n, t, lt[N], rt[N], pos[N];
    
    void modify(int l, int r, ll d){
        int p = pos[l], q = pos[r];
        if(p == q){
            for(int i = l; i <= r; i ++) a[i] += d;
            sum[p] += d * (r - l + 1);
        }
        else{
            for(int i = p + 1; i <= q - 1; i ++) add[i] += d;
            for(int i = l; i <= rt[p]; i ++) a[i] += d;
            sum[p] += d * (rt[p] - l + 1);
            for(int i = lt[q]; i <= r; i ++) a[i] += d;
            sum[q] += d * (r - lt[q] + 1);
        }
    }
    
    ll query(int l, int r){
        int p = pos[l], q = pos[r];
        ll ret = 0;
        if(p == q){
            for(int i = l; i <= r; i ++) ret += a[i];
            ret += add[p] * (r - l + 1);
        }
        else{
            for(int i = p + 1; i <= q - 1; i ++)
                ret += sum[i] + add[i] * (rt[i] - lt[i] + 1);
            for(int i = l; i <= rt[p]; i ++)
                ret += a[i];
            ret += add[p] * (rt[p] - l + 1);
            for(int i = lt[q]; i <= r; i ++)
                ret += a[i];
            ret += add[q] * (r - lt[q] + 1);
        }
        return ret;
    }
    
    
    int main(){
    
        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 ++){
            for(int j = lt[i]; j <= rt[i]; j ++){
                pos[j] = i, sum[i] += a[j];
            }
        }
        for(int i = 1; i <= n; i ++){
            int opt, l, r; ll c;
            cin >> opt >> l >> r >> c;
            if(!opt) modify(l, r, c);
            else cout << query(l, r) % (c + 1) << endl;
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/onionQAQ/p/10876686.html
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