P5534 【XR-3】等差数列

#include <cstdio>
using namespace std;
long long series[100005];
int main()
{
	long long int_a = 0, int_b = 0, n = 0, ans = 0, tolerance = 0;
	scanf ("%d%d%d", &int_a, &int_b, &n);
	tolerance = int_b - int_a;//求出公差 
	series[1] = int_a; series[2] = int_b;//数列前2项 
	for (int i = 3; i <= n; i++)
		series[i] += series[i - 1] + tolerance;//到n的数列的值 
	for (int i = 1; i <= n; i++)
		ans += series[i];//前n项的和
	printf ("%d", ans);
	return 0;
}

各位大佬帮忙看一看，谢谢