1035：等差数列末项计算
时间:2024-05-28 11:16
作者:lizq
点击:次
#include iostream#include iomanip#include cmathusing namespace std;int main(){int a1,a2,n;cina1a2n;int b=a2-a1;couta1+(n-1)*b;return 0;}
#include <iostream>
#include <iomanip>
#include <cmath>
using namespace std;
int main(){
int a1,a2,n;
cin>>a1>>a2>>n;
int b=a2-a1;
cout<<a1+(n-1)*b;
return 0;
}
(责任编辑：lizq) |