URAL 1991 The battle near the swamp 水题

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#数据结构与算法

本文介绍了一道关于计算两个军队战斗后剩余武器数量及未被摧毁怪物数量的问题。文章给出了题目的背景故事、输入输出说明，并提供了一个简单的C++代码实现。

The battle near the swamp

题目连接：

http://acm.timus.ru/problem.aspx?space=1&num=1991

Description

Gungan: Jar Jar, usen da booma!
Jar Jar: What? Mesa no have a booma!
Gungan: Here. Taken dis one.
In the battle with the Trade Federation, Queen Amidala decided to ask gungans for help. Jar Jar Binks escorted the Queen and her people to the holy place where they had an agreement. The gungans agreed to provide their army in order to get the droids of the Federation out from the capital. The gungan ruler Boss Nass was so grateful for uniting the nations that he appointed Jar Jar a general.
And here they are: two armies lined up along the bank of the swamp. The droids of the Federation are well-disciplined soldiers. They stand in neat formation, divided into n blocks of k droids each. The gungans have a foolproof weapon against droids, which is small energy balls called boom booms. One such ball can disable exactly one droid.
Jar Jar Binks also decided to split his army into n parts and give each part a task to destroy the corresponding block of droids. Each part received a truck with boom booms. Now help general Binks calculate the number of boom booms that will be left unused and the number of droids that will survive the attack. You can assume that when a boom boom is fired at a droid by a gungan, it always hits the target.

Input

The first line of the input contains numbers n and k (1 ≤ n, k ≤ 10 000). The second line contains n numbers ai (0 ≤ ai ≤ 100 000) — the number of boom-booms in the i-th truck.

Output

Print two integers — the number of unused boom booms and the number of survived droids.

Sample Input

4 5
2 7 5 0

Sample Output

2 8

Hint

题意

有n块，每块有k个怪物

然后每一块有ai个武器

问你最后剩下多少个武器，以及多少个怪物没有消灭

题解：

水题……

代码

#include<bits/stdc++.h>
using namespace std;

int main()
{
    int n;
    long long k;
    cin>>n>>k;
    long long ans1 = 0,ans2 = 0;
    for(int i=1;i<=n;i++)
    {
        long long x;
        cin>>x;
        if(x>k)ans1+=x-k;
        if(k>x)ans2+=k-x;
    }
    cout<<ans1<<" "<<ans2<<endl;
}