Problem A	Bit Mask
Time Limit	1 Second

In bit-wise expression, mask is a common term. You can get a certain bit-pattern using mask. For example, if you want to make first 4 bits of a 32-bit number zero, you can use 0xFFFFFFF0 as mask and perform a bit-wise AND operation. Here you have to find such a bit-mask.

Consider you are given a 32-bit unsigned integer N. You have to find a mask M such that L ≤ M ≤ U and N OR M is maximum. For example, if N is 100 and L = 50, U = 60 then M will be 59 and N OR M will be 127 which is maximum. If several value of M satisfies the same criteria then you have to print the minimum value of M.

Input
Each input starts with 3 unsigned integers N, L, U where L ≤ U. Input is terminated by EOF.

Output
For each input, print in a line the minimum value of M, which makes N OR M maximum.

Look, a brute force solution may not end within the time limit.

Sample Input	Output for Sample Input
100 50 60 100 50 50 100 0 100 1 0 100 15 1 15	59 50 27 100 1

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如果原本那個 bit 為 1, 選擇與否取決於有沒有大於低限
原本那個 bit 為 0, 選擇與否取決於有沒有小於上限

#include <stdio.h>

int main() {
    long long N, L, R, ans, l, r;
    while(scanf("%lld %lld %lld", &N, &L, &R) == 3) {
        ans = 0;
        long long i;
        for(i = 31; i >= 0; i--) {
            if(N&(1LL<<i)) {
                r = ans + (1LL<<i);
                if(r <= L)
                    ans += 1LL<<i;
            } else {
                l = ans + (1LL<<i);
                if(l <= R)
                    ans += 1LL<<i;
            }
        }
        printf("%lld\n", ans);
    }
    return 0;
}