题解 | #等差数列#
等差数列
https://www.nowcoder.com/practice/f792cb014ed0474fb8f53389e7d9c07f
#include <stdio.h>
int main() {
int n;
while (scanf("%d", &n) != EOF) { // 注意 while 处理多个 case
printf("%d", n*2+(n*(n-1)*3)/2);
}
return 0;
}
