A+B in Hogwarts

Description

If you are a fan of Harry Potter, you would know the school he is attending to, the Hogwarts School for Witchcraft and Wizardry. What you might not know is that Harry is never good at math, yet he doesn't want to bother Hermione for trivial A+B problems. Now your job is to write a simple calculator for him -- the calculator is so simple that it handles non-negative integer additions only.

What is not so simple is that the world of Witchcraft and Wizardry dosen't use a fixed radix numeration system. That is, an integer is represented with different radices for different digits -- the radix for the i-th digit is the i-th prime number. For example, the decimal number 2 is 10 in Hogwarts (let's denote it by 2₁₀ = 1,0_H) since the radix for the 1st digit is the 1st prime number 2. Similarly we have 6₁₀ = 1,0,0_H since the radix for the 2nd digit is 3.

Input

The input consists of multiple test cases, each occupies a line with two integers in Hogwarts' system. Digits of a Hogwarts' integer are seperated by ',' and the two numbers are seperated by a space. Each number has at most 25 digits.

Output

For each test case, print in one line the sum of the two given numbers in Hogwarts' system.

Sample Input

1,0 2,14,2,0 1,2,01 10,6,4,2,1

Sample Output

1,0,11,1,1,01,0,0,0,0,0这道题题意理解就容易做了，题意说的是第i为以第i个素数为基数，6 = 100 = 1*2*3+0*2+0所以这道题只需要按大数的方式处理，进位就可以了。
#include <stdio.h>#include <string.h>#include <stdlib.h>const int maxn = 35, N = 10005;int prime[maxn], vis[N], a[maxn], b[maxn], ans[maxn];char str[N], ch[N];/*第i位就是第i个素数例如100 = 0+0*2+1*2*3 = 6*/void praseInput ( char str[], int a[] ){    char * pos;    int cnt = 0;    while ( pos = strrchr ( str, ',' ) )    //如果成功，则返回指定字符最后一次出现位置的地址，如果失败，则返回 false    {        * pos = '\0';        a[cnt ++] = atoi ( pos+1 ); //从pos+1开始转换数字    }    a[cnt ++] = atoi ( str );}void add ( ){    for ( int i = 0; i < maxn-1; i ++ )    {        ans[i] = ans[i]+a[i]+b[i];  //需要加上本身        ans[i+1] = ans[i]/prime[i];        ans[i] = ans[i]%prime[i];    }}void print ( ){    int i;    for ( i = maxn-1; i >= 0 && ans[i] == 0; i -- )            ;   //从后往前打印,去掉0    while ( i > 0 )    {        printf ( "%d,", ans[i] );        i --;    }    printf ( "%d\n", ans[0] );}void test_print ( int a[] ){    for ( int i = 0; i < maxn; i ++ )        printf ( "%d ", a[i] );    printf ( "\n" );}int main ( ){    int cnt = 0;    vis[0] = vis[1] = 1;    for ( int i = 2; i < N; i ++ )  //求出素数    {        if ( ! vis[i] )            prime[cnt ++] = i;        for ( int j = 0; j < cnt && i*prime[j] < N; j ++ )        {            vis[ i*prime[j] ] = 1;            if ( i%prime[j] == 0 )                break ;        }        if ( cnt > maxn )            break ;    }    while ( ~ scanf ( "%s%s", str, ch ) )    {        memset ( a, 0, sizeof ( a ) );        memset ( b, 0, sizeof ( b ) );        memset ( ans, 0, sizeof ( ans ) );        praseInput ( str, a );        praseInput ( ch, b );        add ( );        print ( );    }    return 0;}