标签：ges prepare osi main constrain single rip putc traints
Our monk loves food. Hence,he took up position of a manager at Sagar,a restaurant that serves people with delicious food packages. It is a very famous place and people are always queuing up to have one of those packages. Each package has a cost associated with it. The packages are kept as a pile. The job of a manager is very difficult. He needs to handle two types of queries:
1) Customer Query:
When a customer demands a package, the food package on the top of the pile is given and the customer is charged according to the cost of the package. This reduces the height of the pile by 1.
In case the pile is empty, the customer goes away empty-handed.
2) Chef Query:
The chef prepares a food package and adds it on top of the pile. And reports the cost of the package to the Manager.
Help him manage the process.
Input:
First line contains an integer Q, the number of queries. Q lines follow.
A Type-1 ( Customer) Query, is indicated by a single integer 1 in the line.
A Type-2 ( Chef) Query, is indicated by two space separated integers 2 and C (cost of the package prepared) .
Output:
For each Type-1 Query, output the price that customer has to pay i.e. cost of the package given to the customer in a new line. If the pile is empty, print "No Food" (without the quotes).
Constraints:
1 ≤ Q ≤ 105
1 ≤ C ≤ 107
用到了slice的两个基本方法 stack = stack[0:stackLength-1] 就是将最后的一个元素删除，slice中下标用的其实是相当[0,stackLength-1} 就是从前标开始，包括前标到后标，但是不包括后标，下来就是往slice中添加元素。
package main
import "fmt"
var stack []int
func main() {
var inputCount int
fmt.Scanln(&inputCount)
stack = make([]int,0,0)
var flag,value int
var stackLength int
for i:=0;i<inputCount;i++{
fmt.Scanln(&flag,&value)
stackLength = len(stack)
if(flag ==1){
if(stackLength == 0){
fmt.Println("No Food")
}else{
fmt.Println(stack[stackLength-1])
stack = stack[0:stackLength-1]
}
}else{
stack = append(stack,value)
}
}
}
标签：ges prepare osi main constrain single rip putc traints
原文地址：http://www.cnblogs.com/jfliuyun/p/6848348.html
