AtCode ABC249 - A - Jogging

• 条件语句、数学

A - Jogging

• https://atcoder.jp/contests/abc249/tasks/abc249_a

Takahashi and Aoki decided to jog.
Takahashi repeats the following: “walk at Bmeters a second for Aseconds and take a rest for Cseconds.”
Aoki repeats the following: “walk at Emeters a second for Dseconds and take a rest for Fseconds.”
When Xseconds have passed since they simultaneously started to jog, which of Takahashi and Aoki goes ahead?

• 1≤A,B,C,D,E,F,X≤100
• All values in input are integers.

Input is given from Standard Input in the following format:

A B C D E F X

When XX seconds have passed since they simultaneously started to jog, if Takahashi goes ahead of Aoki, print Takahashi; if Aoki goes ahead of Takahashi, print Aoki; if they have advanced the same distance, print Draw.

4 3 3 6 2 5 10

Takahashi

During the first 10 seconds after they started to jog, they move as follows.

• Takahashi walks for 4seconds, takes a rest for 3 seconds, and walks again for 3 seconds. As a result, he advances a total of (4+3)×3=21 meters.
• Aoki walks for 6 seconds and takes a rest for 4 seconds. As a result, he advances a total of 6×2=12 meters.

Since Takahashi goes ahead, Takahashi should be printed.

• 比较简单：一个人走A秒，歇C秒，然后再走，X秒后，看谁走的远？

这道题不难，不过相对以往比赛第一题，稍有难度。

我在线上参加比赛时编写的代码如下，是不是写的比较绕了。

主要是当时心态有些急，如果稍加分析，理清思路，很快就能秒掉该题。

void coder_solution() {
    // 提升cin、cout效率
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    // 提升cin、cout效率
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    int a, b, c, d, e, f, x, g, h;
    cin >> a >> b >> c >> d >> e >> f >> x;
    if (x <= a + c) {
        if (x >= a && x <= a + c) {
            g = a * b;
        } else {
            g = x * b;
        }
    }
    if (x <= d + f) {
        if (x >= d && x <= d + f) {
            h = d * e;
        } else {
            h = x * e;
        }
    }
    if (x > a + c) {
        g = x / ( a + c) * a * b;
        if (x % ( a + c) >= a) {
            g += a * b;
        } else {
            g += x % ( a + c) * b;
        }
    }
    if (x > d + f) {
        h = x / ( d + f) * d * e;
        if (x % ( d + f) > d) {
            h += d * e;
        } else {
            h += x % (d + f) * e;
        }
    }
    if ( g > h) {
        cout << "Takahashi";
    } else if ( g == h) {
        cout << "Draw";
    } else {
        cout << "Aoki";
    }
}

• 这个是官方给的题解，利用循环做处理，感觉也有些绕。

void best_solution() {
    // 提升cin、cout效率
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    
    int a, b, c, d, e, f, x;
    cin >> a >> b >> c >> d >> e >> f >> x;
    int takahashi = 0, aoki = 0;
    for (int k = 0; k < x; ++k) {
        if (k % (a + c) < a) {
            takahashi += b;
        }
        if (k % (d + f) < d) {
            aoki += e;
        }
    }
    if (takahashi > aoki) {
        cout << "Takahashin";
    } else if (takahashi < aoki) {
        cout << "Aokin";
    } else {
        cout << "Drawn";
    }
}

这个是今天晚上做复盘时，重写的代码，比之前线上比赛时简洁多了。

主要是还学习到了新知识。（呵呵…)

void best_solution() {
    // 提升cin、cout效率
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    
    // 输入
    int a, b, c, d, e, f, x;
    cin >> a >> b >> c >> d >> e >> f >> x;

    // 定义函数
    auto calc = [&](int a, int b, int c, int x) {
        int m = x / (a + c) * a * b;
        if (x % (a + c) >= a) {
            m += a * b;
        } else {
            m += x % (a + c) * b;
        }
        return m;
    };

    // 分别计算
    int takahashi = calc(a, b, c, x);
    int aoki = calc(d, e, f, x);
    
    // 输出
    if (takahashi > aoki) {
        cout << "Takahashi";
    } else if (takahashi < aoki) {
        cout << "Aoki";
    } else {
        cout << "Draw";
    }
}

• 每次做第一题时，心态都有些急，难免思路就会不清晰。其实精心梳理下，这道题很快就能解决。
• 函数：重复的处理要编写函数，提高程序的扩展性。
  • 老码农叮嘱我：不要为了写代码而写代码，每一次动手敲都要有提高，可以是技术，可以是思维，总之要有提升。

• 关于函数需要深入学习
  • 给老码农安排个活，帮我整理下知识点和学习资料