Solution: analog, with examples to deduce recursion relations: 10--4--1, relationship oldDay = 2 * (newDay + 1).
#include <stdio.h>
int main()
{
int N, newDay, oldDay;
scanf("%d", &N);
newDay = 1; // 第N天剩1个。
while (N != 1)
{ //推导到第1天时,退出循环。
oldDay = (newDay + 1) * 2; //前一天的个数。
newDay = oldDay;
N--;
}
printf("%d\n", oldDay);
return 0;
}