CF-1921-F-根号分治

合集 - Codeforces(31)

1.CF-55-D-数位DP2024-01-232.CF-87-D-并查集2024-01-133.CF-91-B-单调栈+二分2024-01-234.CF-240-F-线段树2024-01-215.CF-242-E-线段树2024-01-206.CF-292-D-并查集2024-01-237.CF-282-E-Trie2024-01-178.CF-342-E-分块2024-01-129.CF-375-D-启发式合并2024-01-1210.CF-431-D-二分+数位DP2024-01-2411.CF-461-B-树形DP2024-01-2212.CF-514-D-单调队列2024-01-1413.CF-522-D-线段树2024-05-0514.CF-570-D-启发式合并2024-01-2115.CF-600-E-启发式合并2024-05-0416.CF-613-D-虚树2024-01-1317.CF-620-E-DFS序+线段树2024-01-1718.CF-675-E-RMQ优化DP2024-01-1619.CF-702-E-倍增2024-01-1720.CF-797-E-根号分治2024-05-0421.CF-817-E-Trie2024-01-1622.CF-877-E-dfs序+线段树2024-05-0523.CF-915-F-并查集2024-01-2124.CF-938-D-最短路2024-05-1025.CF-1051-F-最短路+最小生成树2024-01-1726.CF-1184-E3-最小生成树+倍增+并查集2024-01-2427.CF-1304-E-倍增LCA+思维2024-01-2528.CF-1399-E2-优先队列2024-01-2229.CF-1515-F-思维2024-02-2430.CF-1831-E-卡特兰数+异或哈希+差分2024-01-22

31.CF-1921-F-根号分治2024-01-24

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1921-F 题目大意

有一个长为$n$的序列$a$，有$q$次询问，对于每次询问：

• 给定$s,d,k$，请输出$\sum _{i=1}^{k}i\ast a_{s+(i-1)\ast d}$

---

Solution

根号分治。

• 对于$d\ge \sqrt{n}$的情况，直接暴力计算即可。
• 对于$d\le \sqrt{n}$的情况，这时需要预处理两个数组：$pre,sum$，这里$pre[d][i]$表示公差为$d$，最后一个元素为$a[i]$的数列的前缀和；$sum[d][i]$表示公差为$d$的数列，$a[i]\ast$其项数的前缀和，递推式子如下：

$$pre[d][i+d]=pre[d][i]+a[i]$$

$$sum[d][i+d]=sum[d][i]+a[i]\ast (i/d+1)$$

这时可以$O(1)$的计算出询问结果为$sum[d][s+d\ast k]-sum[d][s]-(pre[d][s+d\ast k]-pre[d][s])\ast (s/d)$。

根号算法结合了暴力计算以及数列求和的优点，真正实现了时空平衡，复杂度均为$O(n\sqrt{n})$。

#include<bits/stdc++.h>
using namespace std;
using ll=long long;

const int N=1e5+10;
ll pre[400][N],sum[400][N];

void solve(){
    int n,q;
    cin>>n>>q;
    int B=sqrt(n);
    vector<ll> a(n+1);
    for(int i=1;i<=n;i++){
        cin>>a[i];
    }
    for(int d=1;d<=B;d++){
        for(int i=1;i<=n;i++){
            pre[d][i+d]=pre[d][i]+a[i];
            sum[d][i+d]=sum[d][i]+a[i]*(i/d+1);
        }
    }
    while(q--){
        int s,d,k;
        cin>>s>>d>>k;
        if(d<B){
            cout<<sum[d][s+d*k]-sum[d][s]-(pre[d][s+d*k]-pre[d][s])*(s/d)<<" ";
        }else{
            ll res=0;
            for(ll i=1;i<=k;i++){
                res+=a[s+d*(i-1)]*i;
            }
            cout<<res<<" ";
        }
    }
    cout<<'\n';
}

int main(){
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    //freopen("input.txt","r",stdin);
    //freopen("output.txt","w",stdout);
    int T=1;
    cin>>T;
    while(T--){
        solve();
    }
    return 0;
}