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锁定老帖子 主题：网上没找到答案的逻辑推理题
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发表时间：2009-09-13
/** * * @author q472732639 * */public class Test { public static void main(String [] args){ TeamObj a = new TeamObj(); //A队 TeamObj b = new TeamObj(); //B队 TeamObj c = new TeamObj(); //C队 int roundNum = 1; //回合数 /* * 下面的纯粹就是按照题意所写的判断 不精湛 粗略的理解为a是必须赢数多.但不一定不输, * b必须是输数少,但不一定去赢,c是积分一定要高,但一定要让A,B两组积分低的情况下不影响战绩..全是废话 */ for(int i = 0 ; i < roundNum ; i ++){ if(a.getWinNum() <= b.getWinNum() || a.getWinNum() <= c.getWinNum()){ a.setWinNum(a.getWinNum() + 1); a.setMin(a.getMin() + 2); if(b.getLoseNum() - c.getLoseNum() > 1){ b.setLoseNum(b.getLoseNum() + 1); }else{ c.setLoseNum(c.getLoseNum() + 1); } }else if(b.getLoseNum() >= c.getLoseNum()){ b.setWinNum(b.getWinNum() + 1); b.setMin(b.getMin() + 2); if(a.getWinNum() - b.getWinNum() > 1){ a.setLoseNum(a.getLoseNum() + 1); }else{ c.setLoseNum(c.getLoseNum() + 1); } }else if(b.getLoseNum() >= a.getLoseNum()){ b.setWinNum(b.getWinNum() + 1); b.setMin(b.getMin() + 2); a.setLoseNum(c.getLoseNum() + 1); }else if(c.getMin() <= a.getMin() || c.getMin() <= b.getMin()){ c.setWinNum(c.getWinNum() + 1); c.setMin(c.getMin() + 2); if(a.getWinNum() - c.getWinNum() > 1){ a.setLoseNum(a.getLoseNum() + 1); }else if(b.getLoseNum() - c.getLoseNum() > 1){ b.setLoseNum(b.getLoseNum() + 1); }else{ c.setDrawNum(b.getDrawNum() + 1); c.setMin(c.getMin() + 1); if(b.getMin() > a.getMin()){ a.setMin(a.getMin() + 1); a.setDrawNum(a.getDrawNum() + 1); }else{ b.setMin(b.getMin() + 1); b.setDrawNum(b.getDrawNum() + 1); } } } if((a.getWinNum() > c.getWinNum() && a.getWinNum() > b.getWinNum()) && (b.getLoseNum() < a.getLoseNum() && b.getLoseNum() < c.getLoseNum() ) && (c.getMin() > a.getMin() && c.getMin() > b.getMin()) ){ System.out.println("a win :" + a.getWinNum() + " lose :" + a.getLoseNum() + " drawnum :" + a.getDrawNum() + " min :" + a.getMin()); System.out.println("b win :" + b.getWinNum() + " lose :" + b.getLoseNum() + " drawnum :" + b.getDrawNum() + " min :" + b.getMin()); System.out.println("c win :" + c.getWinNum() + " lose :" + c.getLoseNum() + " drawnum :" + c.getDrawNum() + " min :" + c.getMin()); System.out.println("roundNum : " + roundNum); break; }else{ System.out.println("a win :" + a.getWinNum() + " lose :" + a.getLoseNum() + " drawnum :" + a.getDrawNum() + " min :" + a.getMin()); System.out.println("b win :" + b.getWinNum() + " lose :" + b.getLoseNum() + " drawnum :" + b.getDrawNum() + " min :" + b.getMin()); System.out.println("c win :" + c.getWinNum() + " lose :" + c.getLoseNum() + " drawnum :" + c.getDrawNum() + " min :" + c.getMin()); System.out.println("-------------------------------------------------------"); roundNum ++ ; } } } } 那个队伍对象就不写了 楼上说的对先用方程式找好数据之间的关系.然后列出来.在解出来就不难了..我这小白代码只是看下.. |
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发表时间：2009-09-13
package filter;
public class TeamObj {
private int min;
private int winNum;
private int loseNum;
private int drawNum;
public int getMin() {
return min;
}
public void setMin(int min) {
this.min = min;
}
public int getWinNum() {
return winNum;
}
public void setWinNum(int winNum) {
this.winNum = winNum;
}
public int getLoseNum() {
return loseNum;
}
public void setLoseNum(int loseNum) {
this.loseNum = loseNum;
}
public int getDrawNum() {
return drawNum;
}
public void setDrawNum(int drawNum) {
this.drawNum = drawNum;
}
}
重新发... |
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发表时间：2009-09-13
/**
*
* @author q472732639
*
*/
public class Test {
public static void main(String [] args){
TeamObj a = new TeamObj(); //A队
TeamObj b = new TeamObj(); //B队
TeamObj c = new TeamObj(); //C队
int roundNum = 1; //回合数
/*
* 下面的纯粹就是按照题意所写的判断 不精湛 粗略的理解为a是必须赢数多.但不一定不输,
* b必须是输数少,但不一定去赢,c是积分一定要高,但一定要让A,B两组积分低的情况下不影响战绩..全是废话
*/
for(int i = 0 ; i < roundNum ; i ++){
if(a.getWinNum() <= b.getWinNum()
|| a.getWinNum() <= c.getWinNum()){
a.setWinNum(a.getWinNum() + 1);
a.setMin(a.getMin() + 2);
if(b.getLoseNum() - c.getLoseNum() > 1){
b.setLoseNum(b.getLoseNum() + 1);
}else{
c.setLoseNum(c.getLoseNum() + 1);
}
}else if(b.getLoseNum() >= c.getLoseNum()){
b.setWinNum(b.getWinNum() + 1);
b.setMin(b.getMin() + 2);
if(a.getWinNum() - b.getWinNum() > 1){
a.setLoseNum(a.getLoseNum() + 1);
}else{
c.setLoseNum(c.getLoseNum() + 1);
}
}else if(b.getLoseNum() >= a.getLoseNum()){
b.setWinNum(b.getWinNum() + 1);
b.setMin(b.getMin() + 2);
a.setLoseNum(c.getLoseNum() + 1);
}else if(c.getMin() <= a.getMin() || c.getMin() <= b.getMin()){
c.setWinNum(c.getWinNum() + 1);
c.setMin(c.getMin() + 2);
if(a.getWinNum() - c.getWinNum() > 1){
a.setLoseNum(a.getLoseNum() + 1);
}else if(b.getLoseNum() - c.getLoseNum() > 1){
b.setLoseNum(b.getLoseNum() + 1);
}else{
c.setDrawNum(b.getDrawNum() + 1);
c.setMin(c.getMin() + 1);
if(b.getMin() > a.getMin()){
a.setMin(a.getMin() + 1);
a.setDrawNum(a.getDrawNum() + 1);
}else{
b.setMin(b.getMin() + 1);
b.setDrawNum(b.getDrawNum() + 1);
}
}
}
if((a.getWinNum() > c.getWinNum() && a.getWinNum() > b.getWinNum())
&& (b.getLoseNum() < a.getLoseNum() && b.getLoseNum() < c.getLoseNum() )
&& (c.getMin() > a.getMin() && c.getMin() > b.getMin()) ){
System.out.println("a win :" + a.getWinNum() + " lose :" + a.getLoseNum() + " drawnum :" + a.getDrawNum() + " min :" + a.getMin());
System.out.println("b win :" + b.getWinNum() + " lose :" + b.getLoseNum() + " drawnum :" + b.getDrawNum() + " min :" + b.getMin());
System.out.println("c win :" + c.getWinNum() + " lose :" + c.getLoseNum() + " drawnum :" + c.getDrawNum() + " min :" + c.getMin());
System.out.println("roundNum : " + roundNum);
break;
}else{
System.out.println("a win :" + a.getWinNum() + " lose :" + a.getLoseNum() + " drawnum :" + a.getDrawNum() + " min :" + a.getMin());
System.out.println("b win :" + b.getWinNum() + " lose :" + b.getLoseNum() + " drawnum :" + b.getDrawNum() + " min :" + b.getMin());
System.out.println("c win :" + c.getWinNum() + " lose :" + c.getLoseNum() + " drawnum :" + c.getDrawNum() + " min :" + c.getMin());
System.out.println("-------------------------------------------------------");
roundNum ++ ;
}
}
}
}
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发表时间：2009-09-13
啊lei lei!..审题没审明白..
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发表时间：2009-09-15
bookong 写道 monsterjiao 写道 如果喜欢足球比赛并关注得的话，心算一下应该就能出来吧：
一个简单的例子就能证明丙有夺冠的可能： 队伍 胜 负 平 总积分 甲 3 5 0 9 乙 1 1 6 8 丙 2 2 4 10 此题解法： 先证明丙会超过乙的积分：乙比并少输，但是输球是0分，而胜是3分，平是1分，那么丙只要将比乙多输的一场放到胜场上，也就是说丙比乙多赢也多输了一场，而平局自然就是乙比丙多了两场，算下来3-2=1，丙超过了乙的积分。 然后证明丙会超过甲的积分：甲比丙多赢一场，光看胜场是多了3分，而甲比并多输的场是可以无限多加入为N吧（反正题目中又没有说三队之前是进行了多少场比较），那么丙的平局的场次自然就比甲多出N-1，（N-1）*1>3的情况很容易出现吧。 希望能帮助到你 对不起，我真没看懂你的例子，我有两个疑问： 1、看看乙吧，1×3+6在什么情况下能等于8？ 2、甲负了5场，乙和丙胜利加起来1+2=3，那两场是跟谁？ 嘿嘿，不好意思；是我的问题；当时重在解题思路了，再说也是思路不够缜密，例子举错了~但也只是错例子而已；谢谢指出。 |
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发表时间：2009-09-15
标准答案其实是这样的：
甲乙两队打平，所以一个“得意洋洋”一个“反唇相讥”。 丙队是垫底的，所以丙队发言“一声不吭”。 |
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