小米OJ-找到第N个数字等差数列的使用

Posted 2021-02-20 一路一歌

 

 

#include<stdio.h>
#include<math.h>
int main(){
  long long n,m,i,x,s;
  int arr[16] = {2,1,2,3,4,5,6,7,8,9,8,7,6,5,4,3};
  scanf("%lld",&n);
  m = sqrt(2*n);///等差数列求和：sn = (a1+an)*n/2;
  for(i = m; i >= 0; i--)
     if(i*(i+1)<2*n)///等差数列
     {
         x = i;///找出上一项
         break;
     }
   s = n - (x+1)*x/2;///当前项的sn减去前一项sn就是当前项
  printf("%d\\n",arr[s%16]);
  return 0;
}