sicily 6484. *the easiest problem

Description
 Have you passed the problem 1000(A - B)? Yeah,it's very easy!
 Now,given two integers N and M(0 < N,M <= 1050 ,N > M),calculate N + M and N * M. I think it's also very easy for you!

Input
 There are several test cases, one line for each case. For each line, there are only two numbers N and M,seperated by space.The input is end by EOF.

Output
Output N + M and N * M in a single line without leading zeros respectively.There is a blank line between two test cases.

用字符串存储数字，运算的时候利用ASCII码改成数字，然后按照小学教的加法和乘法算法来做，先在纸上模拟一遍，然后转化成一堆选择和循环就变成代码了。注意进位的处理。

查了喂鸡才知道乘法原来有那么多奇葩的算法………

 输出格式要求比较奇葩，是下面这样，红色画起来的是输入

#include <stdio.h>
#include <string.h>

void add( const char n[], const char m[] );
void multiply( const char n[], const char m[] );
void printArray( const int array[], int length );
int max( int a, int b);

int main()
{
    char n[55] = {0};
    char m[55] = {0};
    
    if( scanf("%s%s", &n, &m) != EOF )
    {
        add( n, m );
        multiply( n, m );
        
        while( scanf("%s%s", &n, &m) != EOF )
        {
            printf("\n");
            
            add( n, m );
            multiply( n, m );
        }
    }
    
    return 0;
}

void add( const char n[], const char m[] )
{
    int temp;
    int carry;
    int i;
     int ni, mi;
    int nlen, mlen, sumlen;    
    int sum[55] = {0};

    nlen = strlen(n);
    mlen = strlen(m);
    
    for( i = 0; i < max( nlen, mlen ); i++ )
    {

        (nlen - i - 1 >= 0) ? (ni = n[nlen - i - 1] - '0') : (ni = 0);
        (mlen - i - 1 >= 0) ? (mi = m[mlen - i - 1] - '0') : (mi = 0);
        
        temp = sum[i] + ni + mi;
        sum[i] = temp % 10;
        carry = temp / 10; 
        sumlen = i;

        if( carry > 0 )
        {
            sum[i+1] = sum[i+1] + carry;
            sumlen = i+1;
        } 
    }
    
    printArray( sum, sumlen );
    
    return;
}

void multiply( const char n[], const char m[] )
{
    int temp;
    int carry;
    int i1, i2;    
     int i; 
    int j, k;
    int nlen, mlen, prdtlen;
    
    int prdt[110] = {0};
    
    nlen = strlen(n);
    mlen = strlen(m);
    carry = 0;
    
    for( i1 = 0, k = nlen - 1; i1 < nlen; i1++, k-- )
    {
        for( i2 = 0, j = mlen - 1; i2 < mlen; i2++, j-- )
        {
            i = i1 + i2;
            temp = prdt[i] + ( n[k] - '0' ) * ( m[j] - '0' ) + carry;
            prdt[i] = temp % 10;
            carry = temp / 10; 
        }
        
        while( carry > 0 )
        {
            i++;
            prdt[i] = prdt[i] + carry % 10;
            carry = carry / 10;
        }
        
        prdtlen = i;
    }
    
    printArray( prdt, prdtlen );
    
    return;
}

void printArray( const int array[], int length )
{
    int i;
    
    for( i = length; i >= 0; i-- )
    {
        printf( "%d", array[i] );
    }
    
    printf("\n");
}


int max( int a, int b)
{
    if ( a > b )
    {
        return a;
    }
    else
    {
        return b;
    }
}