题解 | #等差数列#
等差数列
https://www.nowcoder.com/practice/f792cb014ed0474fb8f53389e7d9c07f
#include <stdio.h>
int main() {
int n;
scanf("%d",&n);
int re = (4+3*(n-1));
int re1;
if(re%2==0)
{
re1 = re/2*n;
}else if(re%2==1)
{
re1 = re/2*n+n/2;
}
printf("%d",re1);
}
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