UVA 10106

题目大意就是高精度A*B,两个高精度相乘，自然的要用到数组。

解题思路：按照小学乘法，数组a分为各个位数与数组b相乘，然后存在第三二维数组数组c[i]中。最后把c中所有加和。因为我菜，尬了三次，第一次忘记把数组倒置，第二次忘记在第三数组计算，头大了。第三次，由于乘法结果，不会考虑到一个为0时，结果为0.反而输出多个0。wa一次。嗯，A+B会了，这题稍微转换一下，就会ac的。

Ac代码如下：

有点臭有点长，见谅

#include <iostream>#include <cstring>using namespace std; void add(char c[], char a, char b[], int n){int len = strlen(b), temp=0, temp2;int m[3000], sum[3000];memset(sum, 0, sizeof(sum));memset(m, 0, sizeof(m));for (int i=0; i<len; i++) m[i+n] = b[len-i-1]-'0';len += n;for (int i=0; i<len; i++){temp = (sum[i] + m[i] * (a-'0')) / 10;sum[i] = (sum[i] + m[i] * (a-'0')) % 10;sum[i+1] = temp;}if (temp){sum[len] = temp;len++;}for (int i=0; i<len; i++)c[i] = sum[i] + '0';} void Add(char a[], char b[]){int temp, sum[3000], n[3000], m[3000];int max, len1=strlen(a), len2=strlen(b);max = len1>len2?len1:len2;memset(n, 0, sizeof(n));memset(m, 0, sizeof(m));memset(sum, 0, sizeof(sum));for (int i=0; i<len1; i++)m[i] = a[i] - '0';for (int i=0; i<len2; i++)n[i] = b[i] - '0';for (int i=0; i<max; i++){temp = (sum[i] + n[i] + m[i]) / 10;sum[i] = (sum[i] + n[i] + m[i]) % 10;sum[i+1] = temp;} if (temp){sum[max] = temp;max++;}for (int i=0; i<max; i++)a[i] = sum[i] + '0';}int main(){int len1, len2;char a[3000], b[3000], c[1000][1000];while (scanf("%s%s", a, b)!=EOF){if (a[0] == '0' || b[0] == '0')cout << "0" << endl;else{len1 = strlen(a);for (int i=0; i<len1; i++)add(c[i], a[i], b, len1-1-i);for (int i=1; i<len1; i++)Add(c[0], c[i]); len2 = strlen(c[0]);for (int i=len2-1; i>=0; i--)cout << c[0][i];cout << endl;for (int i=0; i<len1; i++)memset(c[i], 0, sizeof(c[i])); } }     return 0;}

嗯...果然代码太臭太长，查了一下网上的代码。get了新方法。一切都在代码中（题解也不给备注，我也不给）

#include <iostream>#include <cstring>using namespace std;void mul(char a[], char b[]){if (a[0] == '0' || b[0] == '0')cout << '0' << endl;else{int len1=strlen(a), len2=strlen(b);int max, m[3000], n[3000], c[3000], temp=0;memset(c, 0, sizeof(c));for (int i=0; i<len1; i++)m[i] = a[len1-i-1]-'0';for (int i=0; i<len2; i++)n[i] = b[len2-i-1]-'0';for (int i=0; i<len1; i++)for (int j=0; j<len2; j++)c[i+j] += m[i] * n[j];max = len1+len2-1;for (int i=0; i<max; i++){temp = c[i] / 10;c[i] = c[i] % 10;c[i+1] += temp;}if (temp)max++;for (int i=max-1; i>=0; i--)cout << c[i];cout << endl;}}int main(){char a[3000], b[3000];while (scanf("%s%s", a, b)!=EOF)mul(a, b);return 0;}