Mail.Ru Cup 2018 Round 1D. Shurikens(思维 + 贪心)

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题目

At a break Vanya came to the class and saw an array of $n$ $k$-bit integers $a_1, a_2, \ldots, a_n$ on the board. An integer $x$ is called a $k$-bit integer if $0 \leq x \leq 2^k - 1$. Of course, Vanya was not able to resist and started changing the numbers written on the board. To ensure that no one will note anything, Vanya allowed himself to make only one type of changes: choose an index of the array $i$ ($1 \leq i \leq n$) and replace the number $a_i$ with the number $\overline{a_i}$. We define $\overline{x}$ for a $k$-bit integer $x$ as the $k$-bit integer such that all its $k$ bits differ from the corresponding bits of $x$. Vanya does not like the number $0$. Therefore, he likes such segments $[l, r]$ ($1 \leq l \leq r \leq n$) such that $a_l \oplus a_{l+1} \oplus \ldots \oplus a_r \neq 0$, where $\oplus$ denotes the bitwise XOR operation. Determine the maximum number of segments he likes Vanya can get applying zero or more operations described above.

中文大意

给我们一个数字$n$和一个数字$k，n$表示数组中的元素个数k表示每个$a_i$的二进制位数不超过$k$我们可以任意对元素操作让$a_i$二进制k位上的数都不同问有多少区间$a_l \oplus a_{l+1} \oplus \ldots \oplus a_r \neq 0$

解法

根据鸽巢原理我们可以知道如果再一段连续的区间异或和中出现了两个相同的值那么就可以确定在这两点之间的区间的异或和是为0的所以本题跟着这个思路来写。
然后我们利用贪心思想从左往右每次让这个区间中连续异或和不为0的数量最大即可

int a[N];
void solve()
{
   int n,k; cin >> n >> k;
   k = 1 << k;
   k -= 1;
   rep(i,n) cin >> a[i];
   map<int,int> mp;
   mp[0] = 1;
   int res = 0;
   for(int i = 1; i <= n; i++) {
       a[i] ^= a[i-1];
       if(mp[a[i]] >= mp[a[i] ^ k]) a[i] ^= k;
       res += mp[a[i]] ++;
   }
   cout << n * (n + 1) / 2 - res << endl;
}