数学/bignum 挑战需要帮助
我正在努力解决以下代码（/挑战），我想知道解决它的最佳方法是什么。
伪（类似）代码
如果我正确理解代码，它会：
var val = 1
foreach (char in firstargument):
val = val * ((ascii)char + 27137)
if (val == 92156295871308407838808214521283596197005567493826981266515267734732800)
print "correct"
else
print "incorrect"
其中“firstargument”是传递给程序的参数，例如：./program 123456...
实际代码
#define _GNU_SOURCE
#include <unistd.h>
#include <string.h>
#include <stdio.h>
#include <gmp.h>
int main(int argc, char *argv[])
{
mpz_t val, mul, cmpval;
char str[513];
int n = 0;
mpz_init(val);
mpz_set_ui(val, 1);
mpz_init(mul);
mpz_init(cmpval);
mpz_set_str(cmpval, "92156295871308407838808214521283596197005567493826981266515267734732800", 10);
if (argc < 2)
{
printf("%s <string>\n", argv[0]);
return -1;
}
strncpy(str, argv[1], 512);
for (n = 0; n < strlen(str); n++)
{
mpz_set_ui(mul, (unsigned long)(str[n] + 27137));
mpz_mul(val, val, mul);
}
if (!(n = mpz_cmp(val, cmpval)))
{
printf("correct.\n");
}
else
{
printf("incorrect.\n");
}
return 0;
}
I'm struggling with the following bit of code(/challenge) and I was wondering what would be the best way to solve it.
Pseudo(-like) code
If I understand the code correctly it does:
var val = 1
foreach (char in firstargument):
val = val * ((ascii)char + 27137)
if (val == 92156295871308407838808214521283596197005567493826981266515267734732800)
print "correct"
else
print "incorrect"
Where 'firstargument' is an argument passed to the program like: ./program 123456...
Actual code
#define _GNU_SOURCE
#include <unistd.h>
#include <string.h>
#include <stdio.h>
#include <gmp.h>
int main(int argc, char *argv[])
{
mpz_t val, mul, cmpval;
char str[513];
int n = 0;
mpz_init(val);
mpz_set_ui(val, 1);
mpz_init(mul);
mpz_init(cmpval);
mpz_set_str(cmpval, "92156295871308407838808214521283596197005567493826981266515267734732800", 10);
if (argc < 2)
{
printf("%s <string>\n", argv[0]);
return -1;
}
strncpy(str, argv[1], 512);
for (n = 0; n < strlen(str); n++)
{
mpz_set_ui(mul, (unsigned long)(str[n] + 27137));
mpz_mul(val, val, mul);
}
if (!(n = mpz_cmp(val, cmpval)))
{
printf("correct.\n");
}
else
{
printf("incorrect.\n");
}
return 0;
}
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评论（3）
我会从大数字必须能被
((ascii)theVeryLastChar + 27137)整除的角度来处理这个问题 - 并尝试找出最后一个字符是什么 - 然后除以它并为“倒数第二个字符”等工作。I would approach this from the point-of-view that the large number must be divisible by
((ascii)theVeryLastChar + 27137)- and try to figure out what this last char is - and then divide by it and work it for the 'the second from last char' etc.这是一个用于计算解决方案的小 Prolog 程序，首先计算具有较低 ASCII 代码的字母。
事实证明只有一种解决方案：
[...]
Here's a little Prolog program to compute the solutions, with the letters with lower ASCII codes first.
It turns out there is only one solution:
[...]
我认为这也被称为中国剩余定理/问题。
第欧根尼算法就是解决方案。
I think this is also known as the chinese remainder theorem/problem.
The diogenes algorithm is the solution.