abc052d

https://atcoder.jp/contests/abc052/tasks/arc067_b

// https://atcoder.jp/contests/abc052/tasks/arc067_b
// 贪心即可, 从左到右行动, 每步选择代价小的方式
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
const int N = 1e5 + 10;
LL n, a, b;
LL x[N];

void solv()
{
    cin >> n >> a >> b;
    LL ans = 0;
    cin >> x[1];
    for (int i = 2; i <= n; i ++)
    {
        cin >> x[i];
        ans += min((x[i] - x[i-1]) * a, b);
    }
    cout << ans << endl;
}

int main()
{
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    int T = 1;
    // cin >> T;
    while (T --)
    {
        solv();
    }
    return 0;
}