Sicily 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 shoshould consist of one line, contain the result of a * B.

Sample Input

12 7

Sample output

14


There is nothing to say: high-precision multiplication. Strings replace numbers for multiplication and convert multiplication to string addition.
Example: "1234" * "123" = "123400" +
"12340" + "12340" +
"1234" + "1234" + "1234"

#include <iostream>#include <string>using namespace std;string add(string a, string b){    if(b.length() > a.length()){        string temp = a;        a = b;        b = temp;    }    a = "0" + a;    int i, len = b.length();    for(i = 0; i < a.length() - len; i++){        b = "0" + b;    }    int in = 0, t;    for(i = a.length() - 1; i >= 0; i--){        t = in;        in = (a[i] - ‘0‘ + b[i] - ‘0‘ + t)/10;        a[i] = (a[i] - ‘0‘ + b[i] - ‘0‘ + t)%10 + ‘0‘;    }    if(a[0] == ‘0‘){        a = a.substr(1, a.length()-1);    }    return a;}string multiply(string a, string b){    string temp;    if(b.length() > a.length()){        temp = a;        a = b;        b = temp;    }    int i, j;    string sum = "0";    for(i = 0; i < b.length(); i++){        temp = a;        for(j = 0; j < b.length()-i-1; j++){            temp = temp + "0";        }        for(j = 0; j < b[i]-‘0‘; j++){            sum = add(sum, temp);        }    }    return sum;}int main(){    int t;    cin >> t;    while(t--){        string a, b;        cin >> a >> b;        cout << multiply(a, b) << endl;    }    return 0;}

 

Sicily 1381. A * B