[Project Euler] Problem 51

Problem Description

By replacing the 1^st digit of *3, it turns out that six of the nine possible values: 13, 23, 43, 53, 73, and 83, are all prime.

By replacing the 3^rd and 4^th digits of 56**3 with the same digit, this 5-digit number is the first example having seven primes among the ten generated numbers, yielding the family: 56003, 56113, 56333, 56443, 56663, 56773, and 56993. Consequently 56003, being the first member of this family, is the smallest prime with this property.

Find the smallest prime which, by replacing part of the number (not necessarily adjacent digits) with the same digit, is part of an eight prime value family.

C++

This problem seems a bit hard to me at the first look. However, through several steps of analysis, I simplify this problem to some extent .

#pragma once

const int MAX_NUM = 1000000;
const int TARGET_COUNT = 8;

int* g_primeArray = NULL;
int g_primeLength = 0;

// pre decalaration
bool CheckThisPrime(int prime, char sameChar, int sameCount);

void Initialize()
{
	g_primeLength = MAX_NUM / 5;
	g_primeArray = new int[g_primeLength];
	MakePrimes(g_primeArray, g_primeLength, MAX_NUM);
}
// By analysis, we can infer that the result should has at least three digits with value 0 or 1 or 2, which are not including the lowest digit of the number.
void Problem_51()
{
	Initialize();

	char digitArray[10] = { 0 };
	int digitCount = 0;
	for(int i =0; i<g_primeLength; i++)
	{
		int curPrime = g_primeArray[i];
		if(curPrime < 1000)
			continue;

		sprintf(digitArray, "%d", curPrime);
		digitCount = strlen(digitArray);

		int zeroCount = 0;
		int oneCount = 0;
		int twoCount = 0;
		for(int j =0; j<digitCount - 1; j++)
		{
			if(digitArray[j] == '0')
				zeroCount++;
			if(digitArray[j] == '1')
				oneCount++;
			if(digitArray[j] == '2')
				twoCount++;
		}

		if(zeroCount < 3 && oneCount < 3 && twoCount < 3)
			continue;

		printf("%d\n", curPrime);

		// int temp = 0;
		// digitArray[0] = '0';
		// sscanf(digitArray, "%d", &temp);
		// printf("%d\n", temp);
		
		if(zeroCount >= 3)
		{
			if(CheckThisPrime(curPrime, '0', zeroCount))
			{
				printf("result = %d\n", curPrime);
				return;
			}
		}
		
		if(oneCount >=3)
		{
			if(CheckThisPrime(curPrime,'1', oneCount))
			{
				printf("result = %d\n", curPrime);
				return;
			}
		}

		if(twoCount >=3)
		{
			if(CheckThisPrime(curPrime, '2', twoCount))
			{
				printf("result = %d\n", curPrime);
				return;
			}
		}

	}
}

bool CheckThisPrime(int prime, char sameChar, int sameCount)
{
	bool ret = false;
	char digitArray[10] = { 0 };
	int digitCount = 0;
	
	sprintf(digitArray, "%d", prime);
	digitCount = strlen(digitArray);

	int* digitIndexArray = new int[sameCount];
	int index=0;
	for(int j =0; j<digitCount - 1; j++)
	{
		if(digitArray[j] == sameChar)
			digitIndexArray[index++] = j;
	}

	// determine which three digits should be modified
	
	bool* useArr = new bool[sameCount];
	useArr[0] = true;
	useArr[1] = true;
	useArr[2] = true;

	while(true)
	{
		char digitArrayBak[10] = { 0 };
		memcpy(digitArrayBak, digitArray, 10 * sizeof(char));
		
		int primeCount = 1;
		int notPrimeCount = 0;
		int notPrimeMax = 2 - (sameChar - '0');
		for(char c=(char)(sameChar + 1); c<='9' ; c++)
		{
			for(int i = 0; i<sameCount; i++)
			{
				if(useArr[i])
				{
					digitArrayBak[digitIndexArray[i]] = c;
				}
			}
			int temp = 0;
			sscanf(digitArrayBak, "%d", &temp);
			if(IsPrime(temp))
			{
				primeCount++;
			}
			else
			{
				notPrimeCount++;
			}

			if(notPrimeCount > notPrimeMax)
			{
				break;
			}

			if(primeCount == TARGET_COUNT)
			{
				ret = true;
				goto LRESULT;
			}
		}

		int reachEndCount = 0;
		for(int i = sameCount -1; i>=0; i--)
		{
			if(useArr[i])
			{
				reachEndCount++;
			}
			else
			{
				break;
			}
		}

		if(reachEndCount == 3)
			break;

		for(int i = sameCount -1 - reachEndCount; i>=0; i--)
		{
			if(useArr[i])
			{
				useArr[i] = false;
				useArr[i+1] = true;
				break;
			}
		}
	}	
	
LRESULT:
	delete[] digitIndexArray;
	delete[] useArr;
	return ret;
}