FOJ 1889 龜兔賽跑

一，問題描述

二，問題分析

思路很簡單，只需要根據題意比較龜兔的路程即可，兔子：（總時間 - 睡覺時間）*速度    烏龜：總時間 * 速度

三，代碼解答

#include<iostream>
#include<cstdio>
#include<vector>
using namespace std;

struct Game
{
	int allTime;		//比賽時間
	int sleepTime;	//兔子睡覺時間
	int rspeed;		//兔子時速
	int tspeed;		//烏龜時速
};

int main() {
	int num;			//比賽組數
	cin >> num;
	vector<Game> games(num);
	vector<int> out(num);
	for (int i = 0; i < num; i++) {
		cin >> games[i].allTime >> games[i].sleepTime >> games[i].rspeed >> games[i].tspeed;
		if ((games[i].allTime - games[i].sleepTime) * games[i].rspeed > games[i].allTime*games[i].tspeed) {		//兔子勝
			out[i] = 0;
		}
		else if ((games[i].allTime - games[i].sleepTime) * games[i].rspeed < games[i].allTime * games[i].tspeed)	//烏龜勝
		{
			out[i] = 1;
		}
		else
		{
			out[i] = -1;
		}
	}
	for (int i = 0; i < num; i++) {
		cout << out[i] << endl;
	}

	return 0;

}