【高精度乘法模板】AOJ-406 Multiplication

Multiplication

Time Limit: 1000 ms   Case Time Limit: 1000 ms   Memory Limit: 64 MB

Description

To calculate the result of two numbers by multiply them

Input

The input consists of multiple test cases.
each line contains two numbers A and B (A,B >= 0) which both are less than 100 digits

Output

Each line contains one result

Sample Input

OriginalTransformed

100 10

Sample Output

OriginalTransformed

1000

————————————————————低落の分割線————————————————————

思路：string是是实现高精度的利器。

代码如下：

/*ID: j.sure.1PROG:LANG: C++*//****************************************/#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <cmath>#include <stack>#include <queue>#include <vector>#include <map>#include <string>#include <climits>#include <iostream>#define LL long longusing namespace std;const int INF = 0x3f3f3f3f;/****************************************/char A[105], B[105];string str[105];void Add(string &s1, string s2) {int i, j, sum = 0;//if(s1.length() < s2.length()) swap(s1, s2);for(i = s1.length()-1, j = s2.length()-1; i >= 0; i--, j--) {   sum += s1[i] - '0';  if(j >= 0) sum += s2[j] - '0'; s1[i] = sum % 10 + '0'; sum /= 10; }if(sum) s1 = '1' + s1;}int main(){#ifdef J_Sure//freopen("000.in", "r", stdin);//freopen(".out", "w", stdout);#endifwhile(~scanf("%s%s", A, B)) {int len_A = strlen(A), len_B = strlen(B);int add, left;for(int i = len_B-1, k = 0; i >= 0, k < len_B; i--, k++) {int _b = B[i] - '0';str[i] = "";add = 0; left = 0;for(int kk = 0; kk < k; kk++)str[i] += '0';for(int j = len_A-1; j >= 0; j--) {int _a = A[j] - '0';int _c = _a * _b + add;add = _c / 10; left = _c % 10;str[i] += left + '0';}if(add)str[i] += add + '0';reverse(str[i].begin(), str[i].end());}for(int i = 1; i < len_B; i++) {Add(str[0], str[i]);}cout << str[0] << endl;}return 0;}