Problem B: ChongBit

Problem B: ChongBit

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 42  Solved: 6
[Submit][Status][Web Board]

Description

Consider you are given a 32-bit unsigned integer N. You have to find a mask M such that L ≤ M ≤ R and N OR M is maximum. For example, if N is 100 and L = 50, R = 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, R where L ≤ R. Input is terminated by EOF.

Output

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

Sample Input

100 50 60

100 50 50

100 0 100

1 0 100

15 1 15

Sample Output

59

50

27

100

1

HINT

分析：

1.对于A|N，对于N取0的二进制位，A全都取1就可以使A|N最大化，可以直接令A为全1;但是，题中要求有多组解时输出最小值，故可令A=~N；

2.题目同时限制了输出结果的取值范围要在[L,R]之内，又要使A|N尽可能大，所以从~N的高位往地位处理，当L,R的二进制值相同时，将~N的该位值也改成这个值，如果L该位为0而R为1，~N的该位值为0时则说明已经满足<R,为1说明已经>L,当都满足过一次时则退出处理。

#include <cstdio>


bool LB[32];
bool RB[32];
bool AB[32];

int main(void)
{
    unsigned N,L,R;
    while (~scanf("%u",&N))
    {
        scanf("%u%u",&L,&R);
        unsigned A=~N;
        for (int i=0;i<32;i++)
        {
            AB[i]=A%2;LB[i]=L%2;RB[i]=R%2;
            A/=2;L/=2;R/=2;
        }
        bool s1=true,s2=true;
        for (int i=31;i>=0&&(s1||s2);i--)
        {
            if (AB[i]>RB[i]&&s2){AB[i]=RB[i];}
            else if (AB[i]<LB[i]&&s1){AB[i]=LB[i];}
            if (AB[i]<RB[i]){s2=false;}
            if (AB[i]>LB[i]){s1=false;}
        }
        unsigned B=0;
        for (int i=31;i>=0;i--)
            {
                B=B<<1;
                B+=AB[i];
            }
        printf("%u\n",B);
    }


    return 0;
}