Codechef Chef and Easy Problem(智商)

Read problems statements in Mandarin chinese, Russian and Vietnamese as well.

You are given a sequence A1, A2, ..., AN and Q queries. In each query, you are given two parameters L and R; you have to find the smallest integer X such that 0 ≤ X < 231and the value of (AL xor X) + (AL+1 xor X) + ... + (AR xor X) is maximum possible.

Note: xor denotes the bitwise xor operation.

Input

• The first line of the input contains two space-separated integers N and Q denoting the number of elements in A and the number of queries respectively.
• The second line contains N space-separated integers A1, A2, ..., AN.
• Each of the next Q lines contains two space-separated integers L and R describing one query.

Output

For each query, print a single line containing one integer — the minimum value of X.

Constraints

• 1 ≤ N ≤ 105
• 1 ≤ Q ≤ 105
• 0 ≤ Ai < 231 for each valid i

Subtasks

Subtask #1 (18 points):

• 1 ≤ N ≤ 103
• 1 ≤ Q ≤ 103
• 0 ≤ Ai < 210 for each valid i

Subtask #2 (82 points): original constraints

Example

Input:

5 3
20 11 18 2 13
1 3
3 5
2 4

Output:

2147483629
2147483645
2147483645

感觉codechef easy里面的都是智商题啊qwq。。

这题我们可以按位考虑。

如果当前为二进制下第$i$位，区间$(l, r)$中元素第$i$位含有$1$的有$x$个

很显然，当$x <= \frac{r - l + 1}{2}$时$0$的个数多，那么答案的第$i$位应为$1$，否则为$0$

#include<cstdio>
#define int long long 
using namespace std;
const int MAXN = 1e5 + 10;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N, Q;
int a[MAXN], sum[32][MAXN];
 main() {
    N = read(); Q = read();
    for(int i = 1; i <= N; i++) a[i] = read();
    for(int j = 0; j <= 30; j++) 
        for(int i = 1; i <= N; i++) 
            sum[j][i] = sum[j][i - 1] + ((a[i] & (1 << j)) != 0);
    while(Q--) {
        int l = read(), r = read(), ans = 0;
        for(int i = 0; i <= 30; i++) 
            if(2 * (sum[i][r] - sum[i][l - 1]) < (r - l + 1))
                ans |= (1 << i);
        printf("%lld\n", ans);
    }
}