构造&贪心题单练习

# 有n个地点，每来往i，j两个地点就花费（i+j）mod（n+1），问遍历所有地点的最低花费
# 当i+j==n+1每两地花费最小，如1, n; 2，n-1
# 只需以1->n->2->n-1->3...这样的顺序花费就最小
n = int(input())

res = 0
# res = (n-1)/2
if n & 1: # n is odd
    res = (n - 1) >> 1
else:
    res = (n >> 1) - 1

print(res)

#include<cstdio>
int main()
{
  int n; scanf("%d",&n);
  printf("%d\n",(n-1)/2);
   return 0;
}