TOYOTA MOTOR CORPORATION Programming Contest 2022(AtCoder Beginner Contest 270) AB题解

A 1-2-4 Test

题意：有1、2、4分值的三道题，三位同学作答，第三位同学回答的题目是第一第二位同学至少一个人回答对的，问第三位同学最大得分。

分析：一开始不好入手，但是看到1，2，4，看到至少一个人回答出来的，就隐约想到了逻辑或运算。这道题就变的特别简单了。

题解：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main()
{int n,m;cin >> n >> m;cout << (m|n);return 0;
}

B Hammer

题意：从坐标0开始，想要去X点，Y点有一面墙，但Z点有把锤子可以砸烂墙。问能不能到达X点，最小需要走多远距离。

分析：感觉这是一个大模拟题，只要枚举出所有的情况，就可以输出结果了。分别是xyz、xzy、yxz、yzx、zyx、zxy，还要判断与0的位置确定正负号。

题解：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main()
{int x,y,z;cin >> x >> y >> z;if(z<y&&y<x){if(x<0){cout << abs(x);}else if(x>0&&y<0){cout << abs(x);}else if(y>0&&z<0){cout << -2*z+x;}else{cout << abs(x);}}else if(z<x&&z<y){if(x>0) cout << abs(x);else if(x<0&&y>0) cout << abs(x);else cout << -1;}else if(y<z&&z<x){if(y>0) cout << -1;if(y<0) cout << abs(x);}else if(y<x&&x<z){if(y>0) cout << -1;if(y<0) cout << abs(x);}else if(x<y&&y<z){if(y>0) cout << abs(x);else if(y<0&&z>0) cout << 2*z-x;else if(y<0&&z<0) cout << abs(x);}else{if(y<0) cout << -1;else cout << abs(x);}return 0;
}

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TOYOTA MOTOR CORPORATION Programming Contest 2022(AtCoder Beginner Contest 270) AB题解

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