A little girl loves problems on bitwise operations very much. Here’s one of them. You are given two integers l and r. Let’s consider the values of for all pairs of integers a and b (l ≤ a ≤ b ≤ r). Your task is to find the maximum value among all considered ones. Expression means applying bitwise excluding or operation to integers x and y. The given operation exists in all modern programming languages, for example, in languages C and Java it is represented as “^”, in Pascal — as «xor». Input The single line contains space-separated integers l and r (1 ≤ l ≤ r ≤ 1018). Please, do not use the %lld specifier to read or write 64-bit integers in С . It is preferred to use the cin, cout streams or the %I64d specifier. Output In a single line print a single integer — the maximum value of for all pairs of integers a, b (l ≤ a ≤ b ≤ r). Example Input 1 2 Output 3 Input 8 16 Output 31 Input 1 1 Output 0
【题解】 首先，异或值要为1，那么两个数的对应二进制位要不同，而异或值要最大，则二进制的高位要尽可能的为1，所以这就是切入点，从给定的区间上下限入手（l和R），从l和r二进制的最高位开始比较，如果出现对应位异或值位1，就从该处开始，低位都置为1，此时其表示的数就是最大异或值了。
代码语言：javascript复制#include<stdio.h>
int main()
{
long long l,r,t,w;
while(scanf("%lld%lld",&l,&r)!=EOF)
{
if(r==l)printf("0n");
else
{w=0;
while(r!=l&&r!=0)//相等或最大值为0 结束 101011
{ //相等则证明 此位之前的异或不可能为1 101110
w ;
l=l>>1;//从右向左移动
r=r>>1;
}
t=1;
while(w--)t=t*2;
printf("%lldn",t-1);
}
}
return 0;
}
