需要7m,11m,17m的管子各10根，至少需要多少根100m的管子，并求出每根管子的利用率
如题：java编码实现！ --------------------编程问答--------------------
import java.math.BigDecimal;
import java.util.ArrayList;
import java.util.List;
/**
* Created by IntelliJ IDEA.
* User: rollet
* Date: 13-1-25
* Time: 下午7:12
* 需要7m,11m,17m的管子各10根，至少需要多少根100m的管子，并求出每根管子的利用率
*/
public class TestTwo {
public static void main(String[] args) {
PIPE test = new PIPE(7, 11, 17, 10, 0, 0, 0, 0);
List<PIPE> list = test.jisuan(100);
for (PIPE pipe : list) {
System.out.print(pipe.getA() + "=" + pipe.getAvalue() + ",");
System.out.print(pipe.getB() + "=" + pipe.getBvalue() + ",");
System.out.print(pipe.getC() + "=" + pipe.getCvalue() + ",");
System.out.println("第" + pipe.getNumvalue() + "根管子使用率为" + new BigDecimal(pipe.getA() * pipe.getAvalue() + pipe.getB() * pipe.getBvalue() + pipe.getC() * pipe.getCvalue()).setScale(0) + "%");
}
}
static class PIPE {
int a;
int b;
int c;
int num;
int avalue;
int bvalue;
int cvalue;
int numvalue;
PIPE(int a, int b, int c, int num, int avalue, int bvalue, int cvalue, int numvalue) {
this.a = a;
this.b = b;
this.c = c;
this.num = num;
this.avalue = avalue;
this.bvalue = bvalue;
this.cvalue = cvalue;
this.numvalue = numvalue;
}
public int getA() {
return a;
}
public int getB() {
return b;
}
public int getC() {
return c;
}
public int getNum() {
return num;
}
public int getAvalue() {
return avalue;
}
public int getBvalue() {
return bvalue;
}
public int getCvalue() {
return cvalue;
}
public int getNumvalue() {
return numvalue;
}
public List<PIPE> jisuan(int i) {
List<PIPE> list = new ArrayList<PIPE>();
int ge = i;
numvalue++;
int alock = 0;
int block = 0;
int clock = 0;
while (ge >= 0) {
if (ge >= c && alock < num) {
ge -= c;
cvalue++;
alock++;
} else if (ge >= b && block < num) {
ge -= b;
bvalue++;
block++;
} else if (ge >= a && clock < num) {
ge -= a;
avalue++;
clock++;
} else {
ge = i;
list.add(new PIPE(a, b, c, num, avalue, bvalue, cvalue, numvalue));
numvalue++;
avalue = 0;
bvalue = 0;
cvalue = 0;
}
if (clock == 10 && alock == 10 && block == 10) {
list.add(new PIPE(a, b, c, num, avalue, bvalue, cvalue, numvalue));
break;
}
}
return list;
}
}
}
--------------------编程问答--------------------
for(int i=100/17;i>0;i--){for(int j=(100-i*17)/11;j>0;j--){
if((100-i*17-j*11)%SEVEN==0){
System.out.println("17厘米的用了="+i);
System.out.println("11厘米的用了="+j);
System.out.println("7厘米的用了="+(100-i*17-j*11)/7);
return ;
}
}
}
不知道是不是你想要的。
补充：Java , Java SE
