Consecutive Sum LightOJ - 1269(区间异或和)

Consecutive Sum

 又来水一发blog...

本来是昨天补codechef的题，最后一道题是可持久化字典树，然后去黄学长博客看了看

觉得字典树写法有点不太一样，就想着用黄学长的板子写码几道题吧...

最后发现...其实是可持久化字典树跟普通字典树写法不一样...我好傻逼...

区间异或和

可以先把前缀异或和插入到字典树中，然后去贪心

求最大值就是要求从高到低尽量不一样

高位不一样优先

用一个bit数组来存放好看多了

偷偷摸摸顺便学了快读板子...虽说可能没啥用...

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

const int maxn = 2e6 + 10;

int sum[50000 + 10];
int ans1, ans2;
int bit[35];
inline int read() {
    int x = 0, f = 1; char ch = getchar();
    while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
    return x * f;
}

struct Trie {
    int tol;
    int Next[maxn][2];
    void init() {
        tol = 0;
        memset(Next, 0, sizeof(Next));
    }
    void insert(int x) {
        int root = 0;
        for (int i = 30; i >= 0; i--) {
            int id = x & bit[i]; id >>= i;
            if (!Next[root][id]) Next[root][id] = ++tol;
            root = Next[root][id];  
        }
    }
    void query(int x) {
        int temp1 = 0, root1 = 0, root2 = 0, temp2 = 0;
        for (int i = 30; i >= 0; i--) {
            int id = x & bit[i]; id >>= i;
            if (Next[root1][id ^ 1]) root1 = Next[root1][id ^ 1], temp1 += bit[i];
            else root1 = Next[root1][id];
            if (Next[root2][id]) root2 = Next[root2][id];
            else root2 = Next[root2][id ^ 1], temp2 += bit[i];
        }
        ans1 = max(ans1, temp1);
        ans2 = min(ans2, temp2);
    }
} trie;

int main() {
    bit[0] = 1; for(int i = 1; i <= 30; i++) bit[i] = bit[i-1] << 1;
    int T;
    T = read();
   // cin >> T;
    int kase = 0; 
    while (T--) {
        int n = read();
      //  Trie trie;
        trie.init();
        ans1 = 0;
        ans2 = 0x3f3f3f3f;
        trie.insert(0);
        for (int i = 1; i <= n; i++) {
            int x; cin >> x;
            sum[i] = sum[i-1] ^ x;
            trie.query(sum[i]);
            trie.insert(sum[i]);
        }
        printf("Case %d: %d %d", ++kase, ans1, ans2);
        puts(""); 
    }
    return 0;
}

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