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

memory limit per test

input

output

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, ..., c_n. The Old Peykan wants to travel from city c₁ to c_n using roads. There are (n - 1) one way roads, the i-th road goes from city c_i to city c_i + 1 and is d_i

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

c_i (except for the last city c_n) has a supply of s_i liters of fuel which immediately transfers to the Old Peykan if it passes the city or stays in it. This supply refreshes instantly k

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.

c_n.

Input

m and k (1 ≤ m, k ≤ 1000). The value m specifies the number of roads between cities which is equal to n - 1.

m space-separated integers d₁, d₂, ..., d_m (1 ≤ d_i ≤ 1000) and the following line contains m space-separated integers s₁, s₂, ..., s_m (1 ≤ s_i ≤ 1000).

Output

c_n 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

c₁

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;
}