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Project Euler 8–Largest product in a series

By Wu Yudong on September 26, 2017

Find the greatest product of five consecutive digits in the 1000-digit number.

73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450

找出以下这个1000位的整数中连续5个数字的最大乘积。(例如前五个数字的乘积是7*3*1*6*7=882)

#include <stdio.h>
#include <stdlib.h>

int main()
{
	FILE *fp;
	char *buffer;
	int i = 0, j = 0;

	fp = fopen("E://file.txt", "r");

	char c;

	while ((c = fgetc(fp))) {
		if (c == EOF)
			break;
		else if (c != '\n')
			i++;
	}
	buffer = (char *)malloc(i * sizeof(char));
	rewind(fp);
	while ((c = fgetc(fp))) {
		if (c == EOF)
			break;
		else if (c != '\n') {
			*(buffer + j) = c;
			j++;
		}
	}
	findmax(buffer, i);
}

int findmax(char *buffer, int i)
{
	int j = 0, max = 0;

	for (j = 0; j < i - 4; j++) {
		if (max < ((buffer[j] - '0') * (buffer[j + 1] - '0') * (buffer[j + 2] - '0') * (buffer[j + 3] - '0') * (buffer[j + 4] - '0')))
			max = ((buffer[j] - '0') * (buffer[j + 1] - '0') * (buffer[j + 2] - '0') * (buffer[j + 3] - '0') * (buffer[j + 4] - '0'));

	}
	printf("\n%d", max);
	return 0;

}

 Answer:40824
Completed on Sun, 17 Nov 2013, 11:16

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