关于"回文数"的编程实验（三）

因为在关于"回文数"的编程实验（二）中有long long数据类型的限制，所以产生的回文数长度有限。

下面的程序尝试采用数组来模拟正整数的加法，这样可以突破语言本身的数据大小限制。

 

另外，本程序还在此基础上，添加了一些和回文数相关的运算，具体功能如下：

 * 直接输入一个正数，判断能否产生回文数

 * 输入 -1 ，程序退出

 * 输入 -2 ，查找长度大于20的回文数

 * 输入 -3，查找在一定范围内需要最多次迭代产生的回文数

 * 输入 -4，统计在一定范围内可以产生回文数的百分比

 

/* * * 任取一个十进制整数，将其倒过来后与原来的整数相加，得到一个新的整数后重复以上步聚， * 则最终可得到一个回文数。请编程验证。 * 回文数的这一形成规则目前还属于一个猜想，尚未得到数学上的证明。 * * 本程序试图使用数组来模拟超大数相加，避免data over flow的错误。 * * 程序支持如下的几种输入： * * 直接输入一个正数，判断能否产生回文数 * 输入 -1 ， 程序退出 * 输入 -2 ， 查找长度大于20的回文数 * 输入 -3， 查找在一定范围内需要最多次迭代产生的回文数 * 输入 -4， 统计在一定范围内可以产生回文数的百分比 * */ #include <stdlib.h> #include <string.h> #include <sys/time.h> #include <iostream> #include <iomanip> using namespace std; #define MAX_DIGITS 64 #define MAX_STEP 100 #define MAX_NUMBER_LENGTH 256 #define MAX_COUNT_TO_CHECK_TIME 50000 #define SEC_TO_REPORT 60 char buffer1[MAX_NUMBER_LENGTH]; char buffer2[MAX_NUMBER_LENGTH]; char buffer3[MAX_NUMBER_LENGTH]; char displayBuf1[MAX_NUMBER_LENGTH]; char displayBuf2[MAX_NUMBER_LENGTH]; char displayBuf3[MAX_NUMBER_LENGTH]; void resetNumBuffers() { for(int i=0; i<MAX_NUMBER_LENGTH;i++ ) { buffer1[i]='0'; buffer2[i]='0'; buffer3[i]='0'; } } // check if this number is a palindrome number bool isPalindrome(char* numBuf, int numLen) { int j; for(j = 0; j < numLen /2; j++) { if (numBuf[MAX_NUMBER_LENGTH - numLen -1 + j] != numBuf[MAX_NUMBER_LENGTH - 2 - j]) { break; } } return (j == numLen/2); } // return a new char buffer representing the resersed num void getReversed(char* numBuf, char* reversedNumBuf, int numLen) { for (int i = 0; i < numLen; i++) { reversedNumBuf[MAX_NUMBER_LENGTH - numLen - 1 + i] = numBuf[MAX_NUMBER_LENGTH - 2 - i]; } reversedNumBuf[MAX_NUMBER_LENGTH -1] = 0; } // return the new sum length int computeSum(char* num1Buf, char* num2Buf, char* sumBuffer, int numLen) { // length of num1 should be equal to num2 // set the sum to all '0' array for(int i=0; i<MAX_NUMBER_LENGTH -1;i++ ) { sumBuffer[i]='0'; } sumBuffer[MAX_NUMBER_LENGTH -1] = 0; // put the 0 ending bool lastNumberGreaterThanTen = false; // now compute to sum from right to left for(int i=0; i<numLen; i++) { int sumPosition = MAX_NUMBER_LENGTH - i - 2; // corresponding sum buffer position int sum2 = num1Buf[sumPosition] - '0' + num2Buf[sumPosition] - '0' + sumBuffer[sumPosition] - '0'; if (sum2 >= 10) { sumBuffer[sumPosition] = '0' + (sum2 % 10); sumBuffer[sumPosition -1]= '1'; lastNumberGreaterThanTen = true; } else { sumBuffer[sumPosition] = '0' + sum2; lastNumberGreaterThanTen = false; } } // check if the sum has more char than the input numbers if (lastNumberGreaterThanTen) { return numLen+1; } else { return numLen; } } // copy the user input number buffer to the internal buffer, right aligned void copyToBuffer(char* inputBuf, char* destBuf, int numLen) { for (int i=0; i< numLen; i++) { destBuf[MAX_NUMBER_LENGTH - numLen - 1 + i] = inputBuf[i]; } destBuf[MAX_NUMBER_LENGTH - 1]=0; // 0 ended } // check if the palindrome number can be generated from input num (within defined max steps) // return length of that number if can, and the steps will contains the step count to generate it // or return 0 if can not int canGeneratePalindrome(char* numBuf, int numLen, bool print = false, int* steps = NULL) { char* currentNumBuf = buffer1; char* reversedNumBuf = buffer2; char* sumNumBuf = buffer3; int currentNumLen = numLen; int newSumNumLen = 0; int round = 1; // copy numBuf to the global buffer copyToBuffer(numBuf, currentNumBuf, numLen); if (print) { cout<<"/n/n"; } while (!isPalindrome(currentNumBuf, currentNumLen)) { getReversed(currentNumBuf, reversedNumBuf, currentNumLen); // get the sum newSumNumLen = computeSum(currentNumBuf, reversedNumBuf, sumNumBuf, currentNumLen); if (print) { strcpy(displayBuf1, (char*)(currentNumBuf + MAX_NUMBER_LENGTH - currentNumLen - 1)); strcpy(displayBuf2, (char*)(reversedNumBuf + MAX_NUMBER_LENGTH - currentNumLen - 1)); strcpy(displayBuf3, (char*)(sumNumBuf + MAX_NUMBER_LENGTH - newSumNumLen - 1)); cout<<setw(3)<<round<<": "; cout<<displayBuf1<<" + "; cout<<displayBuf2<<" = "; cout<<displayBuf3<<endl; } round++; // swap the currentNumBuf and sumNumBuf char* temp = sumNumBuf; sumNumBuf = currentNumBuf; currentNumBuf = temp; currentNumLen = newSumNumLen; if (round > MAX_STEP) { if (print) { cout<<"Failed to generate it!/n/n"; } return 0; } } if (steps != NULL) { *steps = round; } return currentNumLen; } /* * */ int main(int argc, char** argv) { char inputBuffer[MAX_DIGITS]; while(true) { resetNumBuffers(); cout<<"/n/nPlease input a positive number OR:"<<endl; cout<<" -1 to quit/n"; cout<<" -2 to find some big numbers/n"; cout<<" -3 to find the most steps number/n"; cout<<" -4 to do statistics/n"; cout<<" (max char len="<<MAX_DIGITS-1<<" ):"; cin>>inputBuffer; inputBuffer[MAX_DIGITS-1]=(char)0; if (strlen(inputBuffer) == 2 && inputBuffer[0] == '-' && inputBuffer[1] == '1') { break; } else if (strlen(inputBuffer) == 2 && inputBuffer[0] == '-' && inputBuffer[1] == '2') { char buf[15]; // try to find the Palindrome number which is longer than 20 digits for(unsigned int i=1; i<~0; i++) { sprintf(buf, "%d", i); if (canGeneratePalindrome((char*)buf, strlen(buf)) > 20) { canGeneratePalindrome((char*)buf, strlen(buf), true); } } return 0; } else if (strlen(inputBuffer) == 2 && inputBuffer[0] == '-' && inputBuffer[1] == '3') { int upperBound; cout<<"/ninput upper bound: "; cin>>upperBound; char buf[15]; int retSteps = 0, maxSteps = 0, theSeedNumber = 0; struct timeval tv; int count = 0; int lastSecond = 0; // try to find the palindrome number that need the most steps to generate. for(unsigned int i=1; i<=upperBound; i++) { sprintf(buf, "%d", i); if (canGeneratePalindrome((char*)buf, strlen(buf), false, &retSteps) != 0) { if (retSteps > maxSteps) { maxSteps = retSteps; theSeedNumber = i; } } count++; if ( count > MAX_COUNT_TO_CHECK_TIME ) { count = 0; gettimeofday(&tv, NULL); if (tv.tv_sec > lastSecond + SEC_TO_REPORT) { lastSecond = tv.tv_sec; cout<<"==================================================/n"; cout<<"current number: "<<i<<endl; cout<<"current maxStep:"<<maxSteps<<" Number:"<<theSeedNumber<<endl; cout<<"Percentage: "<<((float)i)/((float)upperBound)*100.0f<<"/%/n"; cout<<"==================================================/n"; } } } cout<<"/n/nMost step to generate palindrome between 1 and "<<upperBound<<" are "<<theSeedNumber<<" !"<<endl; sprintf(buf, "%d", theSeedNumber); canGeneratePalindrome((char*)buf, strlen(buf), true); } else if (strlen(inputBuffer) == 2 && inputBuffer[0] == '-' && inputBuffer[1] == '4') { int upperBound; cout<<"/ninput upper bound: "; cin>>upperBound; char buf[15]; int generated = 0, notGenerated = 0; // try to do the statistics within upperBound for(unsigned int i=1; i<=upperBound; i++) { sprintf(buf, "%d", i); if (canGeneratePalindrome((char*)buf, strlen(buf)) != 0) { generated++; } else { notGenerated++; } } cout<<"/n/nResult:"; cout<<"Numbers from 1 to "<<upperBound<<" are checked !"<<endl; cout<<"Generated: "<<generated<<endl; cout<<"Not generated: "<<notGenerated<<endl; cout<<"Percentage:"<<(float)generated/(float)upperBound*100.0f<<"/%/n"<<endl; } else { if (canGeneratePalindrome((char*)inputBuffer, strlen(inputBuffer)) != 0) { canGeneratePalindrome((char*)inputBuffer, strlen(inputBuffer), true); } else { cout<<"Failed to generate Palindrome within "<<MAX_STEP<<" steps !/n/n"; } } } return (EXIT_SUCCESS); }