2006 Zhejiang University Mars a+b

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

Sample output:

1,0,11,1,1,01,0,0,0,0,0

#include <cstdio>#include<cstring>#include<algorithm>using namespacestd;Const intmaxn=1005;BOOLISPRIME[MAXN];intprime[maxn],cnt;voidsieve () {memset (IsPrime,true,sizeof(IsPrime));  for(intI=2; i<= +; i++)    {        if(Isprime[i]) {prime[cnt++]=i;  for(intj=i+i;j<= +; j+=i) isprime[j]=false; }    }}CharSA[MAXN];CharSB[MAXN];structbigint{intE[MAXN]; intLen; BigInt () {memset (E,0,sizeof(e)); Len=0; } BigInt (Const Chars[]) {memset (E,0,sizeof(e)); Len=0; intL=strlen (s); intz=0;  for(intI=0; i<l;i++)        {            if(s[i]==',') {E[len++]=Z; Z=0; }            Else{z*=Ten; Z=z+ (s[i]-'0'); }} E[len++]=Z;  for(intI=0; i<len/2; i++)        {            intt=E[i]; E[i]=e[len-i-1]; E[len-i-1]=T; }} BigIntoperator+(ConstBigInt &t)Const{BigInt res; Res.len=Max (Len,t.len); intup=0; inti;  for(i=0; i<res.len;i++)        {            intz=e[i]+t.e[i]+Up ; Res.e[i]=z%Prime[i]; up=z/Prime[i]; }         while(up!=0) {Res.e[res.len++]=up%Prime[i]; up/=Prime[i]; I++; }        returnRes; }    voidprint () { for(inti=len-1; i>=1; i--) {printf ("%d,", E[i]); } printf ("%d\n", e[0]); }};intMain () {sieve ();  while(SCANF ("%s%s", SA,SB)! =EOF)        {BigInt a (SA);        BigInt b (SB); if((a.len==1&&a.e[0]==0)|| (b.len==1&&b.e[0]==0))             Break;        BigInt Res; Res=a+b;    Res.print (); }    return 0;}/************************************************************** problem:1016 User:banehunter language:c++ result:accepted time:10 Ms memory:1028 kb****************************************************************/

2006 Zhejiang University mars a+b