Codeforces Round #456 (Div. 2) B. New Year's Eve
Since Grisha behaved well last year, at New Year's Eve he was visited by Ded Moroz who brought an enormous bag of gifts with him! The bag contains n sweet candies from the good ol' bakery, each labeled from 1 to n corresponding to its tastiness. No two candies have the same tastiness.
The choice of candies has a direct effect on Grisha's happiness. One can assume that he should take the tastiest ones — but no, the holiday magic turns things upside down. It is the xor-sum of tastinesses that matters, not the ordinary sum!
A xor-sum of a sequence of integers a1, a2, ..., am is defined as the bitwise XOR of all its elements: , here denotes the bitwise XOR operation; more about bitwise XOR can be found here.
Ded Moroz warned Grisha he has more houses to visit, so Grisha can take no more than k candies from the bag. Help Grisha determine the largest xor-sum (largest xor-sum means maximum happiness!) he can obtain.
The sole string contains two integers n and k (1 ≤ k ≤ n ≤ 1018).
Output one number — the largest possible xor-sum.
4 3
7
6 6
7
分析:k>1时,当数字n用二进制表示时,只要把所有数位变成1即可;
k=1直接输出n。
#include<cstdio> int main() { long long N,k; scanf("%lld%lld",&N,&k); if(k==1) printf("%lld\n",N); else { long long ans=1; while(true) { if(ans>N) break; ans=ans*2; } printf("%lld\n",ans-1); } return 0; }
作者:ACRykl —— O ever youthful,O ever weeping!
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