topic:
A, B, C three people and two 2 to 99 integers, C knows these two numbers, he tells the two numbers and tells A, then tell C, then ask A, B: " What is the two numbers? "A said:" I don't know these two numbers, but I can affirm B, I don't know! "I listened to A, I said:" I didn't know, but listened to A. Just know how much it is, "said that" "I also know how much." What is the number of these two? "
I have three groups of answers: 4, 13; 4,61; 16,73
First, because a affirm b does not know, we assume that a known to M, then for equation x y = m, in the integer solution between 2 to 99, there is at least one of the compliance.
Because B listened to a, I will know, set x * y = n, set the integer solution between its 2 to 99 to X1, Y1; X2, Y2; ... xi, yi; for any set of Xi, Yi, finding the sum, and then push once according to the thought of A. If there is only one set of solutions to satisfying a condition, then B can determine that these two numbers are positive solution.
Simultaneously listening to B, then the idea of all the solutions to B. If there is only one set of solutions, then they are correct.
The procedures I wrote are as follows:
#include
#define max 99 # Define min 2 # define true 1 # define false 0
INT MUTEX = 0; // Mutually exclusive semaphore INT FINDOUTFLAG = 0; // When you find a satisfaction, you will set the FindoutFlag 1. INT A = 0, b = 0; // Since the two numbers
// my functionint twokey (int CHECKA (INT M); // a Depending on his condition. INT CHECKB (INT M); // B can be judged according to his conditions. Int isprimenumber (int m); // is not a number of prime
INT TWOKEY (INT M) {IF (m-3 <= (m 1) / 2) {Return False;} Return True;
INT CHECKA (INT M) {INT FLAG = 0; for (int i = min; i <= (m 1) / 2; i ) {IF (Isprimenumber (MI) && isprimenumber (i)) {FLAG = 2; Return false;} if (i == (m 1) / 2 && mutex == 0) {FindOutflag = 0; for (int J = 2; j <= (m 1) / 2; J ) {IF ( Checkb ((mj) * j) == true) {FindOutflag ;} if (Findoutflag> 1) {Break;} if (j == (m 1) / 2 && findoutflag == 1) {IF (a B <= 101) Printf ("% D,% D / N", A, B);}}} f (i == (m 1) / 2) {return true;}}}} int Checkb (int N) {Int getNumber [50] [2]; int group = 0; int ok = 0, Number;
For (int i = min; i <= n / i; i ) {if ((N% i) == 0) {getNumber [group] [0] = i; getNumber [group] [1] = n / i ; Group ;}} mutex = 1; for (int J = group - 1; j> = 0; j -) {if (CHECKA (GetNumber [J] [0] GetNumber [J])! = True) {} else {ok ; number = j;}} if (ok == 1) {a = getNumber [number] [0]; b = getNumber [Number] [1]; mutex = 0; ok--; Return True;} mutex = 0; return false;
}
INT ISPRIMENUMBER (INT M) {for (INT i = min; i <= (m 1) / 2; i ) {IF ((M% I) == 0) {Return False;}} return true;
Void main () {for (INT i = min; i <= 2 * max-1; i ) {if (Twokey (i) == true) Checka (i);}}
