1900思维题

• https://codeforces.com/problemset/problem/1718/A2

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题意：

• 给一个长度为$n$的数组$a$
• 每次操作选择一个区间$[L,R](1 \leq L \ leq R \leq n)$,和一个整数x
• 把区间所有的数异或上x，每次操作的的代价是$（R - L + 1）/2$
• 问把数组$a$全部变为$0$的最小代价是多少

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思路：

• 观察发现$[L,R]$长度为1或2、3或4……的代价是相等的，而且还可以进一步推断，每一次的操作都可以分割成由1或2组成的小区间（在代价上数值相等）
• 于是可见我们要分割的小区间长度为2的要尽可能的多（或者说每次操作都贪心少掉一次），才能达到代价最小
• 令$pre[i] = a[1] $^$ a[2] $^$......$^$a[i]$,对于一个区间[L,R],当pre[L] == pre[R]时，容易发现这个区间操作的最小代价是R-L-1，否则是R-L
• 于是题目就转化成了一个简单的线性dp，时间复杂度O（n^2）的做法呼之欲出
• 但是显然O（n^2）是不太够用的，我们需要优化，用一个map来记录上一个前缀异或和最近的位置，从这个位置进行状态转移即可
• 时间复杂度是O(n)

---

#include<bits/stdc++.h>
#define debug1(a) cout<<#a<<'='<< a << endl;
#define debug2(a,b) cout<<#a<<" = "<<a<<"  "<<#b<<" = "<<b<<endl;
#define debug3(a,b,c) cout<<#a<<" = "<<a<<"  "<<#b<<" = "<<b<<"  "<<#c<<" = "<<c<<endl;
#define debug4(a,b,c,d) cout<<#a<<" = "<<a<<"  "<<#b<<" = "<<b<<"  "<<#c<<" = "<<c<<"  "<<#d<<" = "<<d<<endl;
#define debug5(a,b,c,d,e) cout<<#a<<" = "<<a<<"  "<<#b<<" = "<<b<<"  "<<#c<<" = "<<c<<"  "<<#d<<" = "<<d<<"  "<<#e<<" = "<<e<<endl;
#define fr(t, i, n)for (long long i = t; i < n; i++)
#define endl "\n"
#define fi first
#define se second
//#define int long long
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;
typedef pair<LL,LL> PLL;

//#pragma GCC optimize(3,"Ofast","inline")
//#pragma GCC optimize(2)
const int N = 1e5+10;
int a[N],pre[N],f[N];

void solve() 
{
    int n;cin >> n;
    for(int i = 1; i <= n; i++)
    {
        cin >> a[i];
        pre[i] = pre[i-1] ^ a[i];
    }

    map<int,int> idx;
    idx[0] = 0;　　　　　　　　　　//需要注意，刚开始的位置记录为0，0
    for(int i = 1;i <= n;i ++)
    {
        f[i] = f[i-1] + 1;
        if(idx.count(pre[i]))
        {
            f[i] = min(f[i],f[idx[pre[i]]] + i - idx[pre[i]] - 1);
        }
        idx[pre[i]] = i;
    }
    cout << f[n] << endl;
}

signed main()
{
    
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    
    int T = 1;cin >> T;
    
    while(T--){
        //puts(solve()?"YES":"NO");
        solve();
    }
    return 0;

}
/*

*/

---

• 从这个O（n）的线性dp，能发现每个位置都是找上一个最近的相同异或前缀和的位置，那么我们可以直接省略掉这个dp的过程，直接用map记录
• 答案就是  n - x（[L,R]，pre[L] == pre[R]最多的数量）

#include<bits/stdc++.h>
#define debug1(a) cout<<#a<<'='<< a << endl;
#define debug2(a,b) cout<<#a<<" = "<<a<<"  "<<#b<<" = "<<b<<endl;
#define debug3(a,b,c) cout<<#a<<" = "<<a<<"  "<<#b<<" = "<<b<<"  "<<#c<<" = "<<c<<endl;
#define debug4(a,b,c,d) cout<<#a<<" = "<<a<<"  "<<#b<<" = "<<b<<"  "<<#c<<" = "<<c<<"  "<<#d<<" = "<<d<<endl;
#define debug5(a,b,c,d,e) cout<<#a<<" = "<<a<<"  "<<#b<<" = "<<b<<"  "<<#c<<" = "<<c<<"  "<<#d<<" = "<<d<<"  "<<#e<<" = "<<e<<endl;
#define fr(t, i, n)for (long long i = t; i < n; i++)
#define endl "\n"
#define fi first
#define se second
//#define int long long
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;
typedef pair<LL,LL> PLL;

//#pragma GCC optimize(3,"Ofast","inline")
//#pragma GCC optimize(2)
const int N = 1e5+10;
int a[N],pre[N];

void solve() 
{
    int n;cin >> n;
    for(int i = 1; i <= n; i++)
    {
        cin >> a[i];
        pre[i] = pre[i-1] ^ a[i];
    }

    set<int> st;st.insert(0);
    int ans = n;
    for(int i = 1;i <= n;i ++)
    {
        if(st.count(pre[i]))
        {
            ans --;
            st.clear();
        }
        st.insert(pre[i]);
    }
    cout << ans << endl;
}

signed main()
{
    
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    
    int T = 1;cin >> T;
    
    while(T--){
        //puts(solve()?"YES":"NO");
        solve();
    }
    return 0;

}
/*

*/