2013
12-21

# ACboy needs your help again!

ACboy was kidnapped!!
he miss his mother very much and is very scare now.You can’t image how dark the room he was put into is, so poor :(.
As a smart ACMer, you want to get ACboy out of the monster’s labyrinth.But when you arrive at the gate of the maze, the monste say :" I have heard that you are very clever, but if can’t solve my problems, you will die with ACboy."
The problems of the monster is shown on the wall:
Each problem’s first line is a integer N(the number of commands), and a word "FIFO" or "FILO".(you are very happy because you know "FIFO" stands for "First In First Out", and "FILO" means "First In Last Out").
and the following N lines, each line is "IN M" or "OUT", (M represent a integer).
and the answer of a problem is a passowrd of a door, so if you want to rescue ACboy, answer the problem carefully!

The input contains multiple test cases.
The first line has one integer,represent the number oftest cases.
And the input of each subproblem are described above.

For each command "OUT", you should output a integer depend on the word is "FIFO" or "FILO", or a word "None" if you don’t have any integer.

4
4 FIFO
IN 1
IN 2
OUT
OUT
4 FILO
IN 1
IN 2
OUT
OUT
5 FIFO
IN 1
IN 2
OUT
OUT
OUT
5 FILO
IN 1
IN 2
OUT
IN 3
OUT

1
2
2
1
1
2
None
2
3

/*
* 1702_1.cpp
*
*  Created on: 2013年8月7日
*      为了能有章泽天这样的女朋友而不断努力。。。。
*/

#include <iostream>
#include <stack>
#include <queue>

using namespace std;

int main() {
int t, m;
string str;
cin >> t;
stack<int> p;
queue<int> q;
while (t--) {
/**
* 因为STL中的stack、queue没有专门的clear()函数
* 所以每次都要自己写清空stack、queue的函数
*/
while(!p.empty()){
p.pop();
}

while(!q.empty()){
q.pop();
}
cin >> m >> str;

if (str == "FIFO") {

for (int i = 0; i < m; ++i) {
string cmd;
int n;
cin >> cmd;
if (cmd == "IN") {
cin >> n;
q.push(n);
} else if (cmd == "OUT") {
if (!q.empty()) {
int a = q.front();
q.pop();
cout << a << endl;
} else {
cout << "None" << endl;
}

}
}
} else {
for (int i = 0; i < m; ++i) {
string cmd;
int n;
cin >> cmd;
if (cmd == "IN") {
cin >> n;
p.push(n);
} else if (cmd == "OUT") {
if (!p.empty()) {
int a = p.top();
p.pop();
cout << a << endl;
} else {
cout << "None" << endl;
}

}
}

}

}

}

1. for(int i=1; i<=m; i++){
for(int j=1; j<=n; j++){
dp = dp [j-1] + 1;
if(s1.charAt(i-1) == s3.charAt(i+j-1))
dp = dp[i-1] + 1;
if(s2.charAt(j-1) == s3.charAt(i+j-1))
dp = Math.max(dp [j - 1] + 1, dp );
}
}
这里的代码似乎有点问题？ dp(i)(j) = dp(i)(j-1) + 1;这个例子System.out.println(ils.isInterleave("aa","dbbca", "aadbbcb"));返回的应该是false

2. 约瑟夫也用说这么长……很成熟的一个问题了，分治的方法解起来o(n)就可以了，有兴趣可以看看具体数学的第一章，关于约瑟夫问题推导出了一系列的结论，很漂亮

3. 这道题目的核心一句话是：取还是不取。
如果当前取，则index+1作为参数。如果当前不取，则任用index作为参数。