Thermostat（思维）

题目链接

题目描述：

Vlad came home and found out that someone had reconfigured the old thermostat to the temperature of $a$.

The thermostat can only be set to a temperature from $l$ to $r$ inclusive, the temperature cannot change by less than $x$. Formally, in one operation you can reconfigure the thermostat from temperature $a$ to temperature $b$ if $|a-b|\ge x$ and $l\le b\le r$.

You are given $l$, $r$, $x$, $a$ and $b$. Find the minimum number of operations required to get temperature $b$ from temperature $a$, or say that it is impossible.

输入描述：

The first line of input data contains the single integer $t$ ($1\le t\le {10}^4$) — the number of test cases in the test.

The descriptions of the test cases follow.

The first line of each case contains three integers $l$, $r$ and $x$ ($-{10}^9\le l\le r\le {10}^9$, $1\le x\le {10}^9$) — range of temperature and minimum temperature change.

The second line of each case contains two integers $a$ and $b$ ($l\le a,b\le r$) — the initial and final temperatures.

输出描述：

Output $t$ numbers, each of which is the answer to the corresponding test case. If it is impossible to achieve the temperature $b$, output $-1$, otherwise output the minimum number of operations.

Note:

In the first example, the thermostat is already set up correctly.

In the second example, you can achieve the desired temperature as follows: $4\to 10\to 5$.

In the third example, you can achieve the desired temperature as follows: $3\to 8\to 2\to 7$.

In the fourth test, it is impossible to make any operation.

AC代码：

#include <bits/stdc++.h>
 
using namespace std;
 
typedef long long LL;
 
const int N = 1e5 + 10;
 
int T;
int l, r, x, a, b;
 
void solve()
{
    cin >> l >> r >> x >> a >> b;
 
    int minn = min(a, b); // 找出 a 和 b 中的最小值
    int maxx = max(a, b);
 
    if (a == b) // 如果 a 和 b 相等则不用移动
    {
        cout << 0 << '\n';
        return;
    }
    if (abs(a - b) >= x) // 如果 a 和 b 的差大于等于 x 则直接将 a 移动到 b
    {
        cout << 1 << '\n';
        return;
    }
    if (maxx + x <= r || minn - x >= l) // 若最大值加上 x 的值仍然在 r 的范围内或者最小值减去 x 的值仍然在 l 的范围内，那么就代表可以只用移动两次
    {
        cout << 2 << '\n';
        return;
    }
    if (minn + x > b && minn + x <= r && maxx - x < a && maxx - x >= l) // 若最小值加上 x 的值在 b 和 r 的范围内并且最大值减去 x 的值在 l 和 a 的范围内，则可以先将最小值和最大值加上减去 x，此时两点的距离必大于 x，需要移动三步
    {
        cout << 3 << '\n';
        return;
    }
 
    cout << -1 << '\n'; // 上述几种情况都不满足则代表无法从 a 移动到 b
}
 
int main()
{
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
 
    // scanf("%d", &T);
    cin >> T;
 
    while (T--)
    {
        solve();
    }
 
    return 0;
}