无论如何，我一直在编写projecteuler.com问题的代码，现在是第二次程序有了，无论我做什么尝试修复它，都返回0.我在C编码。
** PROJECT EULER SPOILER ALERT 的 ** * ** * *
有这个问题的第一个程序是找到可以通过乘以两个三位数来产生的最大回文。
/* this is a program to find the largest palindrome produced
by two factors of n digits each */
// test combos of n digit factors for palindrome product
// run backwards to optimize speed for finding largest
// design should work for user input number of digits to be multiplied
#include <stdio.h>
int main()
{
// declare and value vars
int n; // number of digits
int nsub; // for first while loop
int a = 0; // count down 1
int b = 0; // count down 2
int c; // counter
int d; // digit to be moved in palindrome checker algorithm
int e; // translates to true/false for palindrome
int t1; // number to be tested for... (palindromnity? palindromicness?)
int t3 = 0; // just another variable in my palindrome checker
int t4; // aaand another variable
int p; // current leading largest palindrome
// get user input for number of digits
printf("Please input the number of digits (<5 recommended) you \n");
printf("would like in the two factors that will be multiplied \n");
printf("to find the largest palindrome they can produce\n");
scanf("%d", &n);
// translate number of digits to count down start
nsub = n;
while ( nsub > 0 )
{
a = 10 * a + 9;
b = a;
nsub--;
}
int i = a; // random int to avoid 'change in a' problem
// start of finding stuff algorithm
for ( c=0 ; c < i ; c++ )
{
// loop to run through all b vals on one a val
while( b > 0 )
{
t1 = a * b;
// this will allow a check for an any digit palindrome
/* the program will remove the last digit (d) from the
(number that is hopefully a palindrome) and move
it as the first digit in a new number to be compared
to the original. before comparing the numbers with
d moved to the other, it will compare them with d
having been completely removed. this way even and odd
digit palindromes will register correct. */
t4 = t1;
while( t4 > 0 )
{
e = 1;
d = t4 % 10;
t4 = (t4 - d)/10;
if( t4 == t3 )
{
e = 0;
break;
}
t3 = 10 * t3 + d;
if( t4 == t3 )
{
e = 0;
break;
}
}
// e = 0 means it is a palindrome
if(e == 0 && p < t1)
{
p = t1;
// p is the current leading largest palindrome
}
b--;
}
// resets at next a val with equal b val
a--;
b = a;
}
printf("The largest palindrome that is the product\n");
printf("of two numbers of %d digits is: %d\n", n, p);
getchar(); // extra for input
getchar();
return 0;
}
无论我使用多少位数，程序总是说我的答案是0。
遇到此问题的下一个程序是找到1000位数字中五个连续数字的最大乘积。这仍在进行中。
/*
73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450
*/
/* this is a program to find the greatest
product of five consecutive digits in the
1000 digit number above.
*/
#include <stdio.h>
#include <string.h>
int main()
{
char str[1001] = "7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450";
int p; // current product
int g = 0; // current greatest product
int a=5; // counter
while( a != 1001 )
{
str[a]*str[a-1]*str[a-2]*str[a-3]*str[a-4] = p;
if( p > g )
{
g = p;
}
else
{
}
a++;
}
printf("%d is the largest product of five consecutive numbers", g);
getchar();
return 0;
}
谢谢，我已经完成了一百万次并且不知道它可能是什么。
答案 0 :(得分：0)
关于第二个问题：
str[a]*str[a-1]*str[a-2]*str[a-3]*str[a-4] = p;
这项任务似乎相反。此外，str是由字符'0' '1' '2'组成的字符串，不是直接数字，或者至少
'0' != 0
因为'0'是字符数字0，其编码为0x48