URAL1991 The battle near the swamp

最新推荐文章于 2024-12-09 17:21:21 发布

Jelly_acm

最新推荐文章于 2024-12-09 17:21:21 发布

阅读量413

点赞数

分类专栏： ----水题

本文链接：https://blog.csdn.net/Jelly_acm/article/details/70145819

版权

----水题专栏收录该内容

75 篇文章

订阅专栏

The battle near the swamp

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 a _i (0 ≤ a _i ≤ 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.

Example

input	output
4 5 2 7 5 0	2 8

—————————————————————————————————————

题目的意思是给出n个地区，每个地区有ai个机器人，每个地区投放k歌炸弹，每个炸弹可以炸死1个机器人，问最后幸存的机器人数量和剩下的炸弹的数量

#include <iostream>
#include <cstdio>
using namespace std;

int main()
{
    int x,y,a;
    while(~scanf("%d%d",&x,&y))
    {
        int ans1=0,ans2=0;
        for(int i=0;i<x;i++)
        {
            scanf("%d",&a);
            if(y>a) ans1+=y-a;
            else  ans2+=a-y;
        }
        printf("%d %d\n",ans2,ans1);
    }
    return 0;
}