微軟面試題之數字謎題 (轉)

微軟面試題之數字謎題 (轉)[@more@]

設有兩個自然數m,n，2〈=m<=99。 S先生知道這兩數的和s，P先生知道這兩數的積p。他們兩人進行了如下的對話：
S：我知道你不知道這兩個數是什麼，但我也不知道。
P：現在我知道這兩個數了。
S：現在我也知道這兩個數了。
由這些條件，試確定m，n。

因為S知道兩數之和，卻由此推斷P不知道兩個數，所以說兩數之和s一定不能拆分成兩個素數的和，即m,n不可能都是素數，且m,n中不會有大於50的素數，否則的話m*n可以唯一分解，P知道了m,n的積就一定可以知道m,n了。

P從S的言語中能夠判斷出的資訊是：
1。m,n不會全是素數；
2。m,n中不會有大於50的素數；
3。m,n之和不能拆成兩個素數的和；
4。因為S自己也不知道這兩個數是什麼，所以這兩個數的和一定小於99+98，否則S就可以知道這兩個數是什麼了。

滿足以上條件的 s=m+n有以下的可能：

11
17
23
27
29
35
37
41
47
196

然後P根據自己掌握的p=m*n立即算出m,n，這說明p=m*n是具有以下性質的特殊數字：

根據這個特殊的p,當s取上面的那些值的時候，只有一種s的取值使得方程
m+n=s,
m*n=p
在[2,99]內有唯一的整數解。

根據這個性質計算出的p有以下的情況（不妨設m<=n）：

p = 18, s= 11, m = 2, n = 9
p = 24, s= 11, m = 3, n = 8
p = 28, s= 11, m = 4, n = 7
p = 50, s= 27, m = 2, n = 25
p = 52, s= 17, m = 4, n = 13
p = 54, s= 29, m = 2, n = 27
p = 76, s= 23, m = 4, n = 19
p = 92, s= 27, m = 4, n = 23
p = 96, s= 35, m = 3, n = 32
p = 100, s= 29, m = 4, n = 25
p = 110, s= 27, m = 5, n = 22
p = 112, s= 23, m = 7, n = 16
p = 114, s= 41, m = 3, n = 38
p = 124, s= 35, m = 4, n = 31
p = 130, s= 23, m = 10, n = 13
p = 138, s= 29, m = 6, n = 23
p = 140, s= 27, m = 7, n = 20
p = 148, s= 41, m = 4, n = 37
p = 150, s= 35, m = 5, n = 30
p = 152, s= 27, m = 8, n = 19
p = 154, s= 29, m = 7, n = 22
p = 160, s= 37, m = 5, n = 32
p = 162, s= 27, m = 9, n = 18
p = 168, s= 29, m = 8, n = 21
p = 170, s= 27, m = 10, n = 17
p = 172, s= 47, m = 4, n = 43
p = 174, s= 35, m = 6, n = 29
p = 176, s= 27, m = 11, n = 16
p = 182, s= 27, m = 13, n = 14
p = 186, s= 37, m = 6, n = 31
p = 190, s= 29, m = 10, n = 19
p = 196, s= 35, m = 7, n = 28
p = 198, s= 29, m = 11, n = 18
p = 204, s= 29, m = 12, n = 17
p = 208, s= 29, m = 13, n = 16
p = 216, s= 35, m = 8, n = 27
p = 232, s= 37, m = 8, n = 29
p = 234, s= 35, m = 9, n = 26
p = 238, s= 41, m = 7, n = 34
p = 246, s= 47, m = 6, n = 41
p = 250, s= 35, m = 10, n = 25
p = 252, s= 37, m = 9, n = 28
p = 270, s= 37, m = 10, n = 27
p = 276, s= 35, m = 12, n = 23
p = 280, s= 47, m = 7, n = 40
p = 288, s= 41, m = 9, n = 32
p = 294, s= 35, m = 14, n = 21
p = 304, s= 35, m = 16, n = 19
p = 306, s= 35, m = 17, n = 18
p = 310, s= 41, m = 10, n = 31
p = 322, s= 37, m = 14, n = 23
p = 336, s= 37, m = 16, n = 21
p = 340, s= 37, m = 17, n = 20
p = 348, s= 41, m = 12, n = 29
p = 364, s= 41, m = 13, n = 28
p = 370, s= 47, m = 10, n = 37
p = 378, s= 41, m = 14, n = 27
p = 390, s= 41, m = 15, n = 26
p = 396, s= 47, m = 11, n = 36
p = 400, s= 41, m = 16, n = 25
p = 408, s= 41, m = 17, n = 24
p = 414, s= 41, m = 18, n = 23
p = 418, s= 41, m = 19, n = 22
p = 442, s= 47, m = 13, n = 34
p = 462, s= 47, m = 14, n = 33
p = 480, s= 47, m = 15, n = 32
p = 496, s= 47, m = 16, n = 31
p = 510, s= 47, m = 17, n = 30
p = 522, s= 47, m = 18, n = 29
p = 532, s= 47, m = 19, n = 28
p = 540, s= 47, m = 20, n = 27
p = 546, s= 47, m = 21, n = 26
p = 550, s= 47, m = 22, n = 25
p = 552, s= 47, m = 23, n = 24
p = 9604, s= 196, m = 98, n = 98

最後P說出自己已經知道m,n以後，S也說自己知道了m,n，這說明S根據自己手中的兩數之和可以推斷出唯一的m,n來。

因此還要去除上面的情況中重複用到s的情況，得到下面的情況：
p = 52, s = 17, m = 4, n = 13
p = 9604, s = 196, m = 98, n = 98

如果規定了m<>n，則最後的解答就是
 m=4 , n=13

下面是：

#include
#include
#include

const int MAX_N = 99;
const char* OUTPUT_FILE = "result.txt";

int s[MAX_N*2];
int p[MAX_N*MAX_N];
int prim[MAX_N];
int primCounter =0;

ofstream fout( OUTPUT_FILE );

// 計算素數
void calPrim()
{
 bool used[MAX_N];
 int i, p=2;
 bool found = true;
 
 prim[primCounter++] = 2;
 memset( used, false, sizeof( used ) );
 
 while( found ) {
 for( i = p; i < MAX_N; i++ )
 if( i % p == 0 )
 used[i] = true;

  found = false;

 for( i = p; i < MAX_N; i++ )
 if( ! used[i] ) {
 p = i;
 prim[primCounter++] = p;
 found = true;
 break;
 }
 }
}

// 根據條件1過濾
void useCon_1()
{
 int i,j;
 memset(s, 0, sizeof(s)); 
 
 for( i = 0; i < 4; i++ ) s[i] =-1;

 calPrim();

 // S可以肯定P不知道這兩個數是什麼

 for( i = 0; i < primCounter; i++ )
 for( j = i; j < primCounter; j++ ) {
 if( prim[i] + prim[j] < MAX_N * 2 )
 s[ prim[i] + prim[j] ] = -1;
 }

 for( i = 0; i < primCounter; i++ )
 if( prim[i] > MAX_N / 2 ) break;

 for( i--; i < primCounter; i++ )
 for( j = 2; j < MAX_N; j++ )
 s[ prim[i] + j ] = -1;

 // 因為S自己也不知道這兩個數是什麼
 for( i = 98 + 99; i < MAX_N + MAX_N; i++ )
 s[i] = -1;

 fout << "滿足S第一句話的兩數之和" << endl;
 for( i = 0; i < MAX_N * 2; i++ )
 if( s[i] == 0 )
 fout << i << endl;
}

// 根據條件2過濾
void useCon_2()
{
 int i, m, n;
 memset( p, 0, sizeof( p ) );

 for( m = 2; m < MAX_N; m++ )
 for( n = 2; n < MAX_N; n++ ) {
 if( s[m+n] >= 0 ) {
 p[m*n]++;
 }
 }

 fout << "滿足P第一句話的兩數之積:" << endl;

 for( i = 0; i < MAX_N * MAX_N; i++ )
 if( p[i] == 1 || p[i] == 2 ) {
 for( m = 2; m < MAX_N; m++ )
 for( n = m; n < MAX_N; n++ )
 if( m * n == i && s[m + n] >= 0 ) {
 fout << "p = " << i
 << ", s= " << m + n
 << ", m = " << m
 << ", n = " << n << endl;
 s[m+n]++;
 } 
 }
 
}

void useCon_3()
{
 int i, m, n;
 fout << "滿足S第二句話的結果:" << endl;
 
 for( i = 0; i < MAX_N * MAX_N; i++ )
 if( p[i] == 1 || p[i] == 2 ) {
 for( m = 2; m < MAX_N; m++ )
 for( n = m; n < MAX_N; n++ )
 if( m * n == i && s[m + n] == 1 ) {
 fout << "p = " << i
 << ", s = " << m + n
 << ", m = " << m
 << ", n = " << n << endl;
 } 
 }
}

void main()
{ 
 useCon_1();
 useCon_2();
 useCon_3(); 
}