题解 | #等差数列#
等差数列
http://www.nowcoder.com/practice/f792cb014ed0474fb8f53389e7d9c07f
直接用数学公式
an=a1+(n-1)*d
sn=(a1+an)*n/2
#include<bits/stdc++.h> using namespace std; int main() { int a1=2,d=3; int n,an,sum; while(cin>>n) { an=a1+(n-1)*d; sum=(a1+an)*n/2; cout<<sum<<endl; sum=0; } }