123 lines
2.5 KiB
Markdown
123 lines
2.5 KiB
Markdown
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---
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title: 炸弹人游戏
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date: 2022-03-09 00:16:12.441
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updated: 2022-09-05 20:04:13.861
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url: https://hhdxw.top/archives/75
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categories:
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- C&C++
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tags:
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- C&C++
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---
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## 一、题目:
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你只有一枚炸弹,这枚炸弹威力超强(杀伤距离超长,可以消灭杀伤范围内的所有敌人)。请问在哪里放置炸弹可以消灭最多的敌人?最多可以消灭多少敌人?
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## 二、注意:
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灰色底纹表示墙,炸弹不能穿透。
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人脸 表示敌人。
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空白位置表示空地。
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炸弹只能安放到空地上。
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![img](https://yovinchen-1308133012.cos.ap-beijing.myqcloud.com/wps1700.tmp.jpg)
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## 三、代码实现如下:
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```c
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#include <bits/stdc++.h>
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using namespace std;
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int main()
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{
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//0.墙 1.表示敌人 2.空地
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int max = 0,p,q;
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int a[13][13] = { 0,0,0,0,0,0,0,0,0,0,0,0,0,
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0,1,1,2,1,1,1,0,1,1,1,2,0,
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0,0,0,2,0,1,0,1,0,1,0,1,0,
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0,2,2,2,2,2,2,2,0,2,2,1,0,
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0,1,0,2,0,0,0,2,0,1,0,1,0,
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0,1,1,2,1,1,1,2,0,2,1,1,0,
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0,1,0,2,0,1,0,2,0,2,0,0,0,
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0,0,1,2,2,2,1,2,2,2,2,2,0,
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0,1,0,2,0,1,0,0,0,2,0,1,0,
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0,2,2,2,1,0,1,1,1,2,1,1,0,
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0,1,0,2,0,1,0,1,0,2,0,1,0,
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0,1,1,2,1,1,1,0,1,2,1,1,0,
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0,0,0,0,0,0,0,0,0,0,0,0,0,};
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// j代表每一列,i代表每一行
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for(int i = 1;i < 12;i++)
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{
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for(int j = 1;j < 12;j++)
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{
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if(a[i][j] == 2)
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{
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//sum临时计算敌人数量,x,y,分别是横纵坐标
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int x,y,sum=0;
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//向上
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x=i;
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y=j;
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//判断不撞到墙就进入,撞到墙就停止移动
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while(a[x][y]!=0)
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{
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//判断是不是敌人,是,则加一
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if(a[x][y]==1)
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sum++;
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//控制移动炸弹范围
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x--;
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}
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//向下
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x=i;
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y=j;
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while(a[x][y]!=0)
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{
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if(a[x][y]==1)
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sum++;
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x++;
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}
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//向左
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x=i;
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y=j;
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while(a[x][y]!=0)
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{
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if(a[x][y]==1)
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sum++;
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y--;
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}
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//向右
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x=i;
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y=j;
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while(a[x][y]!=0)
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{
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if(a[x][y]==1)
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sum++;
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y++;
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}
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//统计四个方向行走过程中碰到的敌人的数量,并与上一次做比较寻找最大值
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if(sum>max)
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{
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max=sum;
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p=i;
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q=j;
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}
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}
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}
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}
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cout << "最多人数是:" << max << "人" << endl;
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cout << p << "行" << q << "列" << endl;
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return 0;
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}
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```
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## 四、运行结果如下:
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![image-20220309101417055](https://yovinchen-1308133012.cos.ap-beijing.myqcloud.com/image-20220309101417055.png)
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