链接:http://www.acm.cs.ecnu.edu.cn/problem.php?problemid=2108
题意是约瑟夫环问题的嵌套。由于这题的数据多大,导致不得不用公式法,设2^k<=n<2^(k+1),则题解为(n-2^k)+1
这题由于n可以到2^63-1,所以在写函数的时候必须要用unsigned long long,不然会溢出
1 #include <iostream> 2 #include <cstdio> 3 #include <cstdlib> 4 #include <cstring> 5 #include <string> 6 #include <algorithm> 7 using namespace std; 8 9 unsigned long long f(unsigned long long n) 10 { 11 unsigned long long i = 1; 12 while (n >= 2 * i) 13 i *= 2; 14 return (n - i) * 2 + 1; 15 } 16 17 int main() 18 { 19 unsigned long long a, b; 20 while (scanf("%llu%llu", &a, &b) != EOF) 21 { 22 while (b--) 23 { 24 if (a == f(a)) 25 break; 26 a = f(a); 27 } 28 printf("%llu\n", a); 29 } 30 return 0; 31 }