Mars a+b

Description

Read two non-25-bit Mars positive integers a and B, calculate a+b. It is important to note that on Mars, integers are not single-binary, and nth-bit is the nth prime. For example: 10 binary number 2 on Earth, on Mars as "1,0", because the number of Mars single-digit is 2 binary, the Earth's 10 number of 38, on Mars as "1,1,1,0", because the Mars single-digit is 2 binary, 10-digit is 3, the Hundred is 5 binary, thousand is 7 binary ...

Input

The test input contains several test cases, one row for each test case, two Martian positive integers a and B, a comma-separated number of adjacent two-digit numbers for the Mars Integer, and a space interval between a and B. When a or B is 0 o'clock input end, the corresponding result does not output.

Output

Output 1 lines for each test case, that is, the a+b value of the Mars notation.

Sample Input

1,0 2,1 4,2,0 1,2,0 1 10,6,4,2,1 0 0

Sample Output

1,0,1 1,1,1,0 1,0,0,0,0,0 Personal Feelings This problem has been tested many times, forgetting the equal sign and the character array is too small to waste time sourcehttp://acm.hust.edu.cn/vjudge/contest/ View.action?cid=89340#problem/d

#include <stdio.h>#include<math.h>#include<string.h>intPrimeintx) {    intFlag =1;  for(inti =2; I <= sqrt (x); i++)    {        if(x% i = =0) {flag=0;  Break; }    }    returnFlag;}intMain () {intp[ -], top =0, Len1, Len2, num1[ -], num2[ -], ans[ -]; Charstr1[ +], str2[ +];  for(inti =2; Top < -; i++)    {        if(Prime (i)) {P[top++] =i; //printf ("%d\n\n", P[top-1]);        }    }     while(~SCANF ("%s%s", STR1, str2)) {        if(strcmp (STR1,"0") ==0|| strcmp (STR2,"0") ==0)             Break; memset (NUM1,0,sizeof(NUM1)); memset (num2,0,sizeof(num2)); memset (ans,0,sizeof(ans)); Top=0; intK =1, CNT; Len1=strlen (STR1); Len2=strlen (STR2); /*for (int i = 0; i < len1; i++) {printf ("%c:%c\n", Str1[i], str2[i]); } printf ("\ n"); */        //printf ("len1 =%d Len2 =%d\n", len1, len2);         for(inti = len1-1; I >=0; i--)        {            if(Str1[i] = =',') {Top++; K=1; Continue; } Num1[top]+ = (Str1[i]-'0') *K; //printf ("i =%d:num1[%d] =%d\n\n", I, Top, num1[top]);K = k*Ten; } CNT= top+1; //printf ("cnt =%d\n", cnt);top =0; K=1;  for(inti = len2-1; I >=0; i--)        {            if(Str2[i] = =',') {Top++; K=1; Continue; } Num2[top]+ = (Str2[i]-'0') *K; //printf ("i =%d:num2[top] =%d\n\n", I, Top, num2[top]);K = k*Ten; }        if(top+1>CNT) CNT= top+1; //printf ("cnt =%d\n", cnt);         for(inti =0; I < CNT; i++) {Ans[i]+ = Num1[i] +Num2[i];  while(Ans[i] >= p[i] && p[i]! =0) {ans[i+1] + = Ans[i]/P[i]; Ans[i]%=P[i]; }            //printf ("ans[%d] =%d, num1[%d] =%d, num2[%d] =%d\n", I, ans[i], I, num1[i], I, num2[i]);K = i+1; }        if(Ans[k]) printf ("%d,", Ans[k]);  for(inti = cnt-1; i >0; i--) printf ("%d,", Ans[i]); printf ("%d\n", ans[0]); }}

Mars a+b