设有两个自然数m,n，2〈=m<=99。 S先生知道这两数的和s，P先生知道这两数的积p。他们两人进行了如下的对话：
S：我知道你不知道这两个数是什么，但我也不知道。
P：现在我知道这两个数了。
S：现在我也知道这两个数了。
由这些条件，试确定m，n。
因为S知道两数之和，却由此推断P不知道两个数，所以说两数之和s一定不能拆分成两个素数的和，即m,n不可能都是素数，且m,n中不会有大于50的素数，否则的话m*n可以唯一分解，P知道了m,n的积就一定可以知道m,n了。
P从S的言语中能够判断出的信息是：
1。m,n不会全是素数；
2。m,n中不会有大于50的素数；
3。m,n之和不能拆成两个素数的和；
4。因为S自己也不知道这两个数是什么，所以这两个数的和一定小于99+98，否则S就可以知道这两个数是什么了。
满足以上条件的 s=m+n有以下的可能：
11
17
23
27
29
35
37
41
47
196
然后P根据自己掌握的p=m*n立即算出m,n，这说明p=m*n是具有以下性质的特殊数字：
根据这个特殊的p,当s取上面的那些值的时候，只有一种s的取值使得方程
m+n=s,
m*n=p
在[2,99]内有唯一的整数解。
根据这个性质计算出的p有以下的情况（不妨设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
最后P说出自己已经知道m,n以后，S也说自己知道了m,n，这说明S根据自己手中的两数之和可以推断出唯一的m,n来。
因此还要去除上面的情况中重复用到s的情况，得到下面的情况：
p = 52, s = 17, m = 4, n = 13
p = 9604, s = 196, m = 98, n = 98
如果规定了m<>n，则最后的解答就是
m=4 , n=13
下面是程序：
#include <iostream.h>
#include <fstream.h>
#include <string.h>
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 );
// 计算素数
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;
}
}
}
// 根据条件1过滤
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不知道这两个数是什么
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;
// 因为S自己也不知道这两个数是什么
for( i = 98 + 99; i < MAX_N + MAX_N; i++ )
s[i] = -1;
fout << "满足S第一句话的两数之和" << endl;
for( i = 0; i < MAX_N * 2; i++ )
if( s[i] == 0 )
fout << i << endl;
}
// 根据条件2过滤
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 << "满足P第一句话的两数之积:" << 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 << "满足S第二句话的结果:" << 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();
}
