sicily 1381. a*b

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1381. a*b

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

Description

Give two positive integers a and b, please help us calculate a*b.

Input

The first line of the input is a positive integer T. T is the number of test cases followed.

Each test case contain two integer a,b (0<=a<=10^100, 0<=b<=10,000) given in one line.

Output

The output of each test case should consist of one line, contain the result of a*b.

Sample Input

12 7

Sample Output

Problem Source

Algorithm Course Examination 2006

高精度加法思想：

 /*乘法的主要思想是把乘法转化为加法进行运算。请先看下面的等式：12345*4=12345+12345+12345+1234512345*20=123450*212345*24=12345*20+12345*4        12345*123=12345*100+12345*20+12345*3*/#include<iostream>#include<string>using namespace std;string LongAdd(string a, string b) {        // 获取两个字符串长度int al = a.length();int bl = b.length();        // 获取两个字符串中较长的int ma = al > bl ? al : bl;        // 补0for (int i = 0; i < ma - al; i++) { a = "0" + a;}for (int i = 0; i < ma - bl; i++) {    b = "0" + b;}        // 长度相同， 在字符串前补0,防止溢出if (al == bl) {a = "0" + a;b = "0" + b;}        al = a.length();for (int i = al - 1; i >= 0; i--) {int num = (int)(a[i] - '0') + (int)(b[i] - '0');                // 如果大于10，进位if (num >= 10) {a[i] = (char)((num - 10) + '0');int k = i - 1;                        // 如果上一位也需要进位while (((int)(a[k] - '0') + 1) >= 10 && k >= 0 ) { int n = (int)(a[k] - '0') + 1; a[k] = (char)((n - 10) + '0'); k--;}a[k] = (char)((a[k] - '0') + 1 + '0');}else {a[i] = (char)(num + '0');} }         // 去掉我们补0多出来的0 while (a[0] == '0') {a.erase(a.begin()); } return a;}string multiply(string a, string b) {        // 如果被乘数为0，直接返回0if (a == "0") {return "0";}        // 乘数只有一位if (b.size() == 1) {int num = (int)(b[0] - '0');if (num == 0) {return "0";}string sum = a;for (int i = 0; i < num - 1; i++) {sum = LongAdd(sum, a);}return sum;}else {int bl = b.length();string sum;for (int i = 0; i < bl; i++) {                        // 获取第i位数字int time = (int)(b[i] - '0');string tmp = a;                        // 以加法的形式实现乘法for (int j = 0; j < time - 1; j++) {tmp = LongAdd(tmp, a);}                        // 补齐对应位数 如 1000 * 123， 第一位1要补2个0， 2补1一个0 for (int j = 0; j < bl - i - 1; j++) {tmp += "0";}                        // 如果当前位是0， 即number * 0 = 0，不需要理睬。否则，把得到结果相加if (time != 0) {sum = LongAdd(sum, tmp);}}return sum;}}int main() {int t;cin >> t;while (t--) {string a, b;cin >> a >> b;cout << multiply(a, b) << endl;}return 0;}</string></iostream>

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