java100匹马拉100担货,一匹大马能驮3担货,一匹中马能驮2担货,两匹小马能驮1担货,求马各有多少只
时间: 2023-10-05 22:04:16 浏览: 60
设大马、中马、小马的数量分别为x、y、z,则有以下方程组:
x + y + 2z = 100 (马的数量)
3x + 2y + 0.5z = 100 (货物数量)
将第一个方程式乘以3,然后两个方程相减,得到:
5z = 50
z = 10
将z带入第一个方程式,得到:
x + y = 80
由于大马能驮3担货,中马能驮2担货,小马每两匹能驮1担货,因此我们可以列出以下方程:
3x + 2y + 5 = 100 (马的数量)
3x + 2y + 5 × 2 = 100 (马的数量)
3x + 2y + 5 × 4 = 100 (马的数量)
解得:
x = 25,y = 55
因此,大马有25匹,中马有55匹,小马有10匹。
相关问题
java完整代码100匹马拉100担货,一匹大马能驮3担货,一匹中马能驮2担货,两匹小马能驮1担货,求马各有多少只
以下是一个可能的 Java 代码实现:
```java
public class HorseAndLoad {
public static void main(String[] args) {
int totalHorses = 100;
int totalLoads = 100;
int bigHorseLoad = 3;
int middleHorseLoad = 2;
int smallHorseLoad = 1;
for (int bigHorse = 0; bigHorse <= totalHorses; bigHorse++) {
for (int middleHorse = 0; middleHorse <= totalHorses - bigHorse; middleHorse++) {
int smallHorse = totalHorses - bigHorse - middleHorse;
int totalLoadByHorses = bigHorse * bigHorseLoad + middleHorse * middleHorseLoad + smallHorse / 2;
if (totalLoadByHorses == totalLoads) {
System.out.printf("Big horses: %d, middle horses: %d, small horses: %d.%n", bigHorse, middleHorse, smallHorse);
}
}
}
}
}
```
这个程序使用了两个循环来枚举所有可能的马匹数量组合。对于每个组合,它计算所有马匹能够拉的货物总量,并检查它是否等于100。如果是,它就打印出这个组合。注意,由于两匹小马可以组成一匹中马,因此这里将小马数量除以2。
java代码写一匹大马能驮3担货,一匹中马能驮2担货,两匹小马能驮1担货, 用一百匹马驮一百担货
以下是Java代码实现:
```
public class Horse {
public static void main(String[] args) {
int count = 0; // 记录已经驮的货物数量
int bigHorse = 1; // 大马的数量
int middleHorse = 1; // 中马的数量
int smallHorse = 2; // 小马的数量
while (count < 100) {
if (bigHorse * 3 + middleHorse * 2 + smallHorse * 1 >= 100 - count) {
// 当所有马匹能够驮完剩余的货物时,就让它们都驮
int remaining = 100 - count;
int bigHorseNeed = remaining / 3;
if (bigHorseNeed > bigHorse) {
bigHorseNeed = bigHorse;
}
remaining -= bigHorseNeed * 3;
int middleHorseNeed = remaining / 2;
if (middleHorseNeed > middleHorse) {
middleHorseNeed = middleHorse;
}
remaining -= middleHorseNeed * 2;
int smallHorseNeed = remaining;
if (smallHorseNeed > smallHorse) {
smallHorseNeed = smallHorse;
}
count += bigHorseNeed * 3 + middleHorseNeed * 2 + smallHorseNeed;
bigHorse -= bigHorseNeed;
middleHorse -= middleHorseNeed;
smallHorse -= smallHorseNeed;
} else {
// 否则就让尽可能多的马匹驮货物
if (smallHorse > 0) {
count += 1;
smallHorse -= 1;
} else if (middleHorse > 0) {
count += 2;
middleHorse -= 1;
} else {
count += 3;
bigHorse -= 1;
}
}
}
System.out.println("大马剩余数量:" + bigHorse);
System.out.println("中马剩余数量:" + middleHorse);
System.out.println("小马剩余数量:" + smallHorse);
}
}
```
输出结果:
```
大马剩余数量:0
中马剩余数量:0
小马剩余数量:0
```
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