Hangzhou Electric ACM 1230 Mars a+b (binary)

Mars a+b

                               time limit:2000/1000 MS (java/others)      memory limit:65536/32768 K (java/others)
                                            Total Submission (s): 13903     accepted Submission (s): 4810

Problem 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 ... The input test inputs contain 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 outputs 1 rows for each test case, which is the value of the a+b of the Mars notation. Sample input1,0 2,14,2,0 1,2,01 10,6,4,2,10 0 Sample output1,0,11,1,1,01,0,0,0,0,0 The most important thing is to deal with the numbers, that is, how to accept every digit, The binary conversion is actually very simple. Although there is a lot of numbers, but not strings, because there will be two digits or even multiple digits on the high, so the good way is to use an array, one is a number, from low to high, analog conversion.

#include <iostream>#include<string.h>#include<math.h>#include<algorithm>using namespacestd;Const intmaxn= Max;//because the maximum number of two reads is 25 bits.intP[MAXN];voidPrime () {//find the first 150 primesp[0]=2; p[1]=3; intk=2;  for(intI=4; k<maxn;i++){        intflag=0;  for(intj=2; j<=sqrt (i); j + +) {//If a number cannot be divisible by 2 of all books to his square root, he is prime.            if(i%j==0) {flag=1;  Break; }        }        if(!flag) p[k++]=i; }}intMain () {intn,m,i,j,k,t;    Prime (); intA[MAXN],B[MAXN]; intLena,lenb; intx, y; Charop;  while(1) {memset (A,0,sizeof(a));//Initialize A and Bmemset (b,0,sizeof(b)); scanf ("%d", &x);//enter the first digit of aI=0; A[i++]=x; scanf ("%c", &op);//characters after receiving a number         while(op==','){//If the character is a space, it indicates that the number after is B.scanf"%d",&x); A[i++]=x; scanf ("%c",&op); } Lena=i; scanf ("%d",&x); I=0; B[i++]=x; scanf ("%c",&op);  while(op==',') {scanf ("%d",&x); B[i++]=x; scanf ("%c",&op); } LenB=i; Std::reverse (A,a+lena);//The number of flashbacks is received, so the order in turn is calculatedStd::reverse (b,b+LenB); if(a[0]==0&&b[0]==0&&lena==1&&lenb==1) Break;//if both A and B are 0, end the program        intmaxnlen=Max (LENA,LENB); intSUM[MAXN]; memset (SUM,0,sizeof(sum));  for(i=0, k=0; i<maxnlen;i++) {//Add A and Bsum[k++]=a[i]+B[i]; }        intlensum=K;  for(i=0; i<lensum+Ten; i++) {//converting A and B calculated numbers into a binary conversion            intCur=sum[i]%p[i];//Current Value            intMod=sum[i]/p[i];//Roundingsum[i]=cur; Sum[i+1]+=MoD; }         for(i=maxn-1; i>=0;-I.)if(sum[i]!=0) Break;//from the very end, that is, the highest bit value starts to output, so to remove the excess 0;j=i; printf ("%d", Sum[j]);  for(i=j-1; i>=0;--i) {printf (",%d", Sum[i]); } printf ("\ n"); }}

Hangzhou Electric ACM 1230 Mars a+b (binary)