小图灵AC 云斗AC，但洛谷上的数据WA#8 95pts

#include <cmath>
#include <iostream>
using namespace std;
int N;
double d;
struct st {
    long long pos, pri;
} p[100005];
int main() {
    cin >> N >> d;
    p[1].pos = 0;
    for (int i = 2; i <= N; i++) {
        long long dis;
        cin >> dis;
        p[i].pos = p[i - 1].pos + dis;
    }
    for (int i = 1; i <= N; i++) cin >> p[i].pri;
    long long ans = 0;
    int last = 1;
    double remain = 0;
    for (int i = 2; i <= N; i++) {
        if (p[i].pri < p[last].pri || i == N) {
            long long dis = p[i].pos - p[last].pos;
            double cost = dis * 1.0 / d;  // 到下一站要花费的油
            if (remain < cost) {
                long long buy = ceil(cost - remain);  // 要买的油
                ans += buy * p[last].pri;             // 买油的钱
                remain += buy * 1.0;
            }
            remain -= cost;
            last = i;
        }
    }
    cout << ans << endl;
    return 0;
}