AtCode ABC249 - A - Jogging
AtCode ABC249 - A - Jogging
标签
- 分支、数学
题目地址
A - Jogging
- https://atcoder.jp/contests/abc249/tasks/abc249_a
问题描述
Problem Statement
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?
Constraints
- 1≤A,B,C,D,E,F,X≤100
- All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B C D E F X
Output
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.
Sample Input 1
4 3 3 6 2 5 10
Sample Output 1
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.
题意
- 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？
思路
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。
题解
小码匠
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";
}
}
官方题解
- 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 << "Takahashi\n";
} else if (takahashi < aoki) {
cout << "Aoki\n";
} else {
cout << "Draw\n";
}
}
小码匠二次解
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";
}
}
复盘
心态：
- 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。
待补知识点
- 关于函数需要深入学习
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