/
Euler008.c
90 lines (78 loc) · 3.3 KB
/
Euler008.c
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
/*
Problem 8 - Largest product in a series
Published on Friday, 11th January 2002, 06:00 pm; Solved by 168676
The four adjacent digits in the 1000-digit number that have the greatest product are 9 ~ 9 ~ 8 ~ 9 = 5832.
73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
125406987471585238630507156932909632952274430557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450
Find the thirteen adjacent digits in the 1000-digit number that have the greatest product. What is the value of this product?
NOtes:
Lesson 1 learned: check to see if the value is bigger than the biggest signed int
- Largest signed int is 2,147,483,647 or (2^31)-1 or 2.1 * 10^9 (2 followed by 9 zeros)
- Largest unsigned int is 4,294,967,295 or (2^32)-1 or 4.2 * 10^9 (4 followed by 9 zeros)
Lesson 2 learned: if you're code is too cluttered, it's probably overcomplicated. You have extra variables and functions.
Lesson 3 learned: To convert from a char numerical value to an int numerical value, just use this: ( <char> - '0' )
*/
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main(void) {
char numberblock[] =
"73167176531330624919225119674426574742355349194934\
96983520312774506326239578318016984801869478851843\
85861560789112949495459501737958331952853208805511\
12540698747158523863050715693290963295227443043557\
66896648950445244523161731856403098711121722383113\
62229893423380308135336276614282806444486645238749\
30358907296290491560440772390713810515859307960866\
70172427121883998797908792274921901699720888093776\
65727333001053367881220235421809751254540594752243\
52584907711670556013604839586446706324415722155397\
53697817977846174064955149290862569321978468622482\
83972241375657056057490261407972968652414535100474\
82166370484403199890008895243450658541227588666881\
16427171479924442928230863465674813919123162824586\
17866458359124566529476545682848912883142607690042\
24219022671055626321111109370544217506941658960408\
07198403850962455444362981230987879927244284909188\
84580156166097919133875499200524063689912560717606\
05886116467109405077541002256983155200055935729725\
71636269561882670428252483600823257530420752963450";
int index;
double sum;
double largest = 0;
int i;
for (index = 0; index <= 1000-13; index++)
{
sum = 1;
for (i = index; i < index + 13;i++)
{
sum *= (double)(numberblock[i]-'0');
if (sum == 0) break;
}
if (sum>largest)
{
printf("%f > %f\n",sum, largest);
largest=sum;
}
}
printf("%f\n", largest);
return 0;
}