c - 输出数量大于应有的数量 - gcc 编译

Question

我的任务是从一个巨大的字符数组中找到两个五位数的数字中找到最大的产品。

你必须通过蛮力来完成。

数组中最大的 5 位数字是 99890，所以最大的乘积就是它自身的乘积，即 9978012100，但我似乎无法得到那个答案，事实上我得到的答案比我运行时的那个还要大程序。

当我更改程序以查找最大的 4 位数字乘积时，我的程序可以运行，但当我查找 5 位数字时它就会崩溃。我想知道这些数字是否对我糟糕的 IDE 或其他东西来说太大了。

有人可以在 gcc 中编译并运行以下代码并告诉我他们得到的答案是什么吗？将不胜感激。

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

int main (void){

char array[1001] = "73167176531330624919225119674426574742355349194934"
"96983520312774506326239578318016984801869478851843"
"85861560789112949495459501737958331952853208805511"
"12540698747158523863050715693290963295227443043557"
"66896648950445244523161731856403098711121722383113"
"62229893423380308135336276614282806444486645238749"
"30358907296290491560440772390713810515859307960866"
"70172427121883998797908792274921901699720888093776"
"65727333001053367881220235421809751254540594752243"
"52584907711670556013604839586446706324415722155397"
"53697817977846174064955149290862569321978468622482"
"83972241375657056057490261407972968652414535100474"
"82166370484403199890008895243450658541227588666881"
"16427171479924442928230863465674813919123162824586"
"17866458359124566529476545682848912883142607690042"
"24219022671055626321111109370544217506941658960408"
"07198403850962455444362981230987879927244284909188"
"84580156166097919133875499200524063689912560717606"
"05886116467109405077541002256983155200055935729725"
"71636269561882670428252483600823257530420752963450";

char *a_ptr, *b_ptr;
char string[6];
int i, j;
int aNumber, bNumber;
long long greatestProd, prodHolder;

a_ptr = &array[0];
b_ptr = &array[0];

for (i=0; i<996; i++){
    a_ptr = &array[i];
    strncpy(string, a_ptr, 5);
    aNumber = atoi(string);

    for (j=0; j<996; j++){
        b_ptr = &array[j];
        strncpy(string, b_ptr, 5);
        bNumber = atoi(string);


        prodHolder = aNumber * bNumber;
        if (prodHolder > greatestProd){
            greatestProd = prodHolder;
        }

    }

}
printf("%d\n", greatestProd);


return 0;
}

Answer 1

prodHolder = (long long)aNumber * (long long)bNumber;