楼上的你怎么算的???
这是我算的要5售票员
public class Test{
public static void main(String[] args) {
int buyer = 100, time = 420;
int busman,batch,spareTime,resultTime;
for (busman=1;busman<10;busman++){
if(buyer%busman!=0){
batch = buyer/busman;
}
else
batch = buyer/busman-1;
if(batch%busman!=0){
batch = batch/busman+1;
}
else
batch = batch/busman;
spareTime = 120-15*busman;
resultTime = batch*spareTime+120;
if(resultTime<time){
System.out.println("busman number:"+busman);
break;
}
}
}
}
这是我算的要5售票员
public class Test{
public static void main(String[] args) {
int buyer = 100, time = 420;
int busman,batch,spareTime,resultTime;
for (busman=1;busman<10;busman++){
if(buyer%busman!=0){
batch = buyer/busman;
}
else
batch = buyer/busman-1;
if(batch%busman!=0){
batch = batch/busman+1;
}
else
batch = batch/busman;
spareTime = 120-15*busman;
resultTime = batch*spareTime+120;
if(resultTime<time){
System.out.println("busman number:"+busman);
break;
}
}
}
}
解决方案 »
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public class Test{
public static void main(String[] args) {
int buyer = 100, time = 420;
int busman,batch,spareTime,resultTime;
for (busman=1;busman<10;busman++){
if(buyer%busman!=0){
batch = buyer/busman;
}
else
batch = buyer/busman-1;
spareTime = 120-15*busman;
resultTime = batch*spareTime+120;
System.out.println("有售票员"+busman+"人时,用户买票时间最长为:"+resultTime+"秒");
if(resultTime<time){
System.out.println("需要售票员:"+busman);
break;
}
}
}
}
5*60/(120-15*(售票员数)>=100/售票员数 售票员数>=20/3时间就是和km3(北落师门) 的算法一样的,不过这个算法只是和售票员数<=8的时候