Codeforces Bayan 2012-2013 Elimination Round / 241A Old Peykan (贪心)

A. Old Peykan
http://codeforces.com/contest/241/problem/A

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

There are n cities in the country where the Old Peykan lives. These cities are located on a straight line, we'll denote them from left to right as c1, c2, ..., cn.
The Old Peykan wants to travel from city c1 to cn using
roads. There are (n - 1) one way roads, the i-th
road goes from city ci to
city ci + 1 and
is di kilometers
long.

The Old Peykan travels 1 kilometer in 1 hour and consumes 1 liter of fuel during this time.

Each city ci (except
for the last city cn)
has a supply of si liters
of fuel which immediately transfers to the Old Peykan if it passes the city or stays in it. This supply refreshes instantly k hours after it transfers.
The Old Peykan can stay in a city for a while and fill its fuel tank many times.

Initially (at time zero) the Old Peykan is at city c1 and s1 liters
of fuel is transferred to it's empty tank from c1's
supply. The Old Peykan's fuel tank capacity is unlimited. Old Peykan can not continue its travel if its tank is emptied strictly between two cities.

Find the minimum time the Old Peykan needs to reach city cn.

Input

The first line of the input contains two space-separated integers m and k (1 ≤ m, k ≤ 1000).
The value m specifies the number of roads between cities which is equal to n - 1.

The next line contains m space-separated integers d1, d2, ..., dm (1 ≤ di ≤ 1000) and
the following line contains m space-separated integers s1, s2, ..., sm (1 ≤ si ≤ 1000).

Output

In the only line of the output print a single integer — the minimum time required for The Old Peykan to reach city cn from
city c1.

Sample test(s)

input

4 6
1 2 5 2
2 3 3 4

output

10

input

2 3
5 6
5 5

output

14

Note

In the second sample above, the Old Peykan stays in c1 for
3 hours.

贪心思路：在s[i]最大的城市加最多的油，详见代码。

完整代码：

/*30ms,0KB*/

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;

int d[1001], s[1001];

int main(void)
{
	int m, k, count = 0, oil = 0, temp, maxn = 0;
	scanf("%d%d", &m, &k);
	for (int i = 0; i < m; ++i)
	{
		scanf("%d", &d[i]);
		count += d[i];
	}
	for (int i = 0; i < m; ++i)
		scanf("%d", &s[i]);
	for (int i = 0; i < m; ++i)
	{
		oil += s[i];
		maxn = max(maxn, s[i]);
		if (d[i] > oil)
		{
			temp = ceil((double)(d[i] - oil) / maxn);
			count += k * temp;
			oil += maxn * temp;
		}
		oil -= d[i];
	}
	printf("%d", count);
	return 0;
}