10106 - Product

  对于大整数的四则运算，需运用数组的相关处理方法来解决，因为现行一般的系统均不超过32位。

  一道Uva online上的题目，几经波折，略见成效，现将代码复制如下，还望赐教。。。。。

 题目出处：uva.onlinejudge.org/index.php

#include < iostream >

 #include < cstring > 

#include < cmath >

using namespace std;

 

const int MAX = 600; //整形数据不会超过600百位数

char BigInt1[MAX];

char BigInt2[MAX];

int mul[MAX]; //存放最后结果

int Big1[MAX];

int Big2[MAX];

 

int main() {

    int m, n, i, len1, len2;

    while (cin.getline(BigInt1, sizeof(BigInt1)) &&

        cin.getline(BigInt2, sizeof(BigInt2))) {

        len1 = strlen(BigInt1);

        len2 = strlen(BigInt2);

 

        memset(Big1, 0, sizeof(Big1));

        memset(Big2, 0, sizeof(Big2));

        memset(mul, 0, sizeof(mul));

        for (i = 0; i < len1; i++)

            Big1[i] = BigInt1[len1 - i - 1] - '0'; //将大整形数据逆序存放以便进位时的操作

        for (i = 0; i < len2; i++)

            Big2[i] = BigInt2[len2 - i - 1] - '0';

 

        for (m = 0; m < len1; m++) {

            int carry = 0; //进位标志

            int t = m;

            for (n = 0; n < MAX; n++) {

                int multi = Big2[n] * Big1[m] + mul[t] + carry;

                mul[t] = multi % 10;

                carry = multi / 10;

                t++;

            }

        }

 

        for (n = MAX - 1; n >= 0; n--)

            if (mul[n])

                break;

 

        if (n >= 0) //计算的结果不为0时

        {

            for (m = n; m >= 0; m--)

                cout << mul[m];

            cout << endl;

        } else //结果为0的特殊情况需输出

            cout << 0 << endl;

    }

    return 0;

}