题解分享
题解分享简介
等差数列末项计算 - 题解
```
//（n(a1 + an)）/ 2
//d = a2 - a1
//an = a1 + (n - 1)*d
#include <bits/stdc++.h>
using namespace std;
int main()
{
int a1,a2,n;
cin >> a1 >> a2 >> n;
int d = a2 - a1;
int an = a1 + (n - 1) * d;
cout << an;
return 0;
}
```
查看全文
0
0
0
0
