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加减乘除24

lzmagic — Tue, 21 Oct 2008 00:13:00 GMT

摘要: 1/**//** 2 * 程序：加减乘除24 3 * 输入：四个1到13整数（输入四个0结束） 4 * 输出：加减乘除得到24的式子，否... 阅读全文

lzmagic 2008-10-21 08:13 发表评论

出栈次序问题

lzmagic — Sun, 27 Apr 2008 06:56:00 GMT

Description

栈是常用的一种数据结构，有n个元素在栈顶端一侧等待进栈，栈顶端另一侧是出栈序列。你已经知道栈的操作有两种：push和pop，前者是将一个元素进栈，后者是将栈顶元素弹出。现在要使用这两种操作，由一个操作序列可以得到一系列的输出序列。请你编程求出对于给定的n，计算并输出由操作数序列1，2，…，n，经过一系列操作可能得到的输出序列总数。

Input

就一个数n(0

Output

一个数，即可能输出序列的总数目。

Sample Input

Sample Output

Solution 1st:

a[i][j] renotes the possible number of the previous j sequences with n = i

(we set a[k][0] = 1 (0 < k <= n), which implies NULL is thought of one possible sequence)

we init that a[1][1] = a[1][0] = 1;

yeild:

a[i][0] = a[i - 1][0];

a[i][j] = a[i][j - 1] + a[i - 1][j]; (1 <= j <= i - 1)

a[i][i] = a[i][i - 1];

the answer wo want to get is a[n][n].

 1#include<iostream>
 2using namespace std;
 3
 4#define MAXN 20
 5
 6int main()
 7{
 8    // init
 9    int a[MAXN][MAXN];
10    a[1][1] = a[1][0] = 1;
11
12    // input the total number pushed into the stack(0 < n && n < 20)
13    int n, i, j;
14    cin >> n;
15
16    // run
17    for(i = 2; i <= n; i++)
18    {
19        a[i][0] = a[i - 1][0];
20        for(j = 1; j <= i - 1; j++)
21        {
22            a[i][j] = a[i][j - 1] + a[i - 1][j];
23        }
24        a[i][j] = a[i][j - 1];
25    }
26
27    // ouput the number of possible sequences poped from the stack
28    cout << a[n][n] << endl;
29    
30    return 0;
31}

Solution 2nd:

denote a[n] the number of possible sequences poped from the stack,

denote i the number of the set si poped before the first number pushed '1' poped,

and denote i the number of the set sj poped after the first number pushed '1' poped,

then we can easily get the conlusion that no number in seti is bigger than any number in setj,

we can depart the sequence into two part and caculate them respectively,

hence the possible number is a[i] * a[j] (0 <= i <= n - 1), i.e.

      a[n] = a[0]*a[n - 1] + a[1]*a[n - 2] + …… + a[n - 2]*a[1] + a[n - 1]*a[0]


 1#include
 2using namespace std;
 3
 4#define MAXN 20
 5
 6int main()
 7{
 8    int a[MAXN];
 9    a[0] = 1;
10    int n, i, j;
11    cin >> n;
12    for(i = 1; i <= n; i++)
13    {
14        for(a[i] = 0, j = 0; j < i; j++)
15            a[i] += a[j] * a[i - 1 - j];
16    }
17    for(i = 1; i <= n; i++)
18        cout << "a[" << i << "] = " << a[i] << endl;
19    return 0;
20}


Solution 3id:

This is Catalan Number :
      a[n] = a[0]*a[n - 1] + a[1]*a[n - 2] + …… + a[n - 2]*a[1] + a[n - 1]*a[0]
      a[n] = C(n , 2 * n) / (n + 1)

lzmagic 2008-04-27 14:56 发表评论

Who's winner?

lzmagic — Sun, 30 Mar 2008 14:49:00 GMT

第一届广东省大学生程序设计大赛暨第六届中山大学程序设计竞赛（2003年）1.7
Nice and Sunsan play the game of multiplication by ultiplying an integer p by one of the number2 2 to 9.Nic always starts with p = 1, does his multiplication.Then Susan multiplies the number, then Nic and so on. Before a game starts, they draw an integer 1<= n <= 4,294,967,295 and the winner is who first reaches p>=n.

Inuput: Each line of input contains one integer number n and ends while n = 0.
Output: For each line of input output one line either
            Nic wins.
            or
            Susan wins.
            Assume that both of them play perfectly.
Sample Input:
162
17
34012226
0
Output for the Sample Input:
Nic wins.
Susan wins.
Nic wins.

1/*Sulution: Nic(1st) wins over [1, 9], Susan(2nd) wins over [10, 18], Nic(1st) wins over [19, 162], Susan(2nd) wins over [163, 344], think a little and easy know why, then wo can get that 9 * 2 -> 18 * 9 -> 162 ->344……
2*/
3
4#include<iostream>
5
6using namespace std;
7
8int main()
9{
10    int n, a, b;
11    bool isFirWin;
12
13    cin >> n;
14    while(n != 0)
15    {
16        a = 1;
17        b = 9;
18        isFirWin = true;
19
20        do
21        {
22            if(a <= n && n <= b) break;
23            else
24            {
25                a = b + 1;
26
27                if(isFirWin)
28                {
29                    b = b * 2;
30                    isFirWin = false;
31                }
32                else
33                {
34                    b = b * 9;
35                    isFirWin = true;
36                }
37            }
38        }while(1);
39
40        if(isFirWin)
41            cout << "Nic wins.\n";
42        else
43            cout << "Susan wins.\n";
44
45        cin >> n;
46    }
47
48    return 0;
49}

lzmagic 2008-03-30 22:49 发表评论