Grab the root (ii) (HDU 4554 rebellious xiaoming hdu 1002 A + B problem II, the conversion of numbers (inversion), the addition of large numbers ... ）

Reversal of Numbers:

is to save the numbers backwards. (*^__^*) hehe ...

General idea: Take the number one one out, there is an array inside, and then turn it into a number, output.

See the code.

1   while (a)           // Take each digit out and take it out 2         {3             num1[i]=a%;   // take each of you out there in the array, realize the reversal of the number 4             i++;            // changes to the array 5             a/=;           6         }

Example of the strike: Hdu 4554 rebellious Xiao Ming

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=4554

The rebellious Xiao Ming

Time limit:3000/1000 MS (java/others) Memory limit:65535/32768 K (java/others)
Total submission (s): 818 Accepted Submission (s): 568

Problem description Rebellious period of Xiao Ming all like to do the opposite, even look at the number is the same (minus), such as:
Xiao Ming would think of 1234 as 4321, 1234 as 4321, 230 as 032 (032=32), 230 as-032 ( -032=-32).

Now, Xiaoming has done some a+b and a-b topics (A, A is an integer and does not contain a leading 0), if you give the correct answers to these questions, can you guess what the answer will be for xiaoming?

Input inputs The first behavior of a positive integer T (t<=10000), indicating that Xiaoming did a total of T-questions.
Next T line, each line is two integers x, y ( -1000000<=x, y<=1000000), x represents the correct answer for A+b, and y represents the correct answer for a-B.
The input is guaranteed to be legal and does not need to consider A or B is a decimal condition.

Output outputs a total of T-lines, each line outputs two integers s t, separated by a space, where s represents the a+b answer that Xiaoming will get, t means that xiaoming will get a A-B answer.

Sample Input320 7-100-140 Output38 247 7-19-23 topic: Chinese title, clear ~ See the code.

1#include <iostream>2#include <cstdio>3#include <cstring>4 5 using namespacestd;6 7 intnum1[Ten],num2[Ten];8 9 intMain ()Ten { One     intT; Ascanf"%d",&T); -      while(t--) -     { the         intX,y,a,b,i=0, j=0; -         intp=0, q=0; -scanf"%d%d",&x,&y); -memset (NUM1,0,sizeof(NUM1)); +memset (num2,0,sizeof(num2)); -A= (x+y)/2; +b=x-A; A          while(a) at         { -num1[i]=a%Ten; -i++; -A/=Ten; -         } -          while(b) in         { -num2[j]=b%Ten; toJ + +; +B/=Ten; -         } the          for(intk=0; k<i; k++)//Change to Numbers *p=num1[k]+p*Ten; $          for(intL=0; l<j; l++)Panax Notoginsengq=num2[l]+q*Ten; -printf ("%d%d\n", p+q,p-q); the     } +     return 0; A}

Large number of additions:

Number is large, if directly add. It's easy to timeout! So the other way is to directly take out a plus, but can not be replaced by numbers.

1 intl=l1>l2?l1:l2;//take one of the longest between two2      for(intI=0; i<l; i++)3     {4Num3[i]=num1[i]+num2[i];//One plus5         if(num3[i-1]>9&&i>=1)//consider rounding, if more than 9 is required .6         {7num3[i]++; 8         }9 Ten     } One     if(num[l-1]>9)//the last one's rounding problem A     { -num[l]++;//still need rounding . -l++;//The length needs to be added one the}

Example: HDU 1002 A + B problem II

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=1002

A + B Problem II

Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 238717 Accepted Submission (s): 46010

Problem Descriptioni has a very simple problem for you. Given integers A and B, your job is to calculate the Sum of A + B.

Inputthe first line of the input contains an integer T (1<=t<=20) which means the number of test cases. Then T lines follow, each line consists of both positive integers, A and B. Notice that the integers is very large, that M EANs should not process them by using 32-bit integer. Assume the length of each integer would not exceed 1000.

Outputfor Each test case, you should output of the lines. The first line was "Case #:", # means the number of the the test case. The second line is the equation "A + b = sum", sum means the result of A + B. Note there is some spaces int the Equati On. Output a blank line between the test cases.

Sample Input21 2112233445566778899 998877665544332211

Sample outputcase 1:1 + 2 = 3Case 2:1.,122,334,455,667,79e,+17 + 998877665544332211 = 1111111111111111110 Refer to the above explanations. Take the number one one out and add it upside down, in the output ~

1#include <iostream>2#include <cstdio>3#include <cstring>4 5 using namespacestd;6 7 intMain ()8 {9     intn,flag=1;Ten     Chara[8000],b[8000]; One     intnum1[8000],num2[8000],ans[8000]; Ascanf"%d",&n); -      while(n--) -     { the         intk=0, s=0; -scanf"%s%s", A, b); -printf ("Case %d:\n", flag++); -         intlen1=strlen (a); +         intLen2=strlen (b); -memset (NUM1,0,sizeof(NUM1)); +memset (num2,0,sizeof(num2)); Amemset (ans,0,sizeof(ans)); at         intL=len1>len2?Len1:len2; -          for(intI=len1-1; i>=0; i--) -         { -num1[k]=a[i]-'0'; -k++; -         } in          for(intj=len2-1; j>=0; j--) -         { tonum2[s]=b[j]-'0'; +s++; -         } the          for(intI=0; i<l; i++) *         { $ans[i]=num1[i]+Num2[i];Panax Notoginseng             if(i>=1) -                 if(ans[i-1]>9) the                 { +ans[i]++; A                 } the                 //cout<<num1[i]<< "" <<num2[i]<< "" <<ans[i]<<endl; +         } -         if(ans[l-1]>9) $         { $ans[l]=1; -l++; -         } theprintf ("%s +%s =", A, b); -          for(inti=l-1; i>=0; i--)Wuyiprintf"%d", ans[i]%Ten); the         if(n) -printf ("\ n"); Wuprintf ("\ n"); -     } About     return 0; $}

Grab the root (ii) (HDU 4554 rebellious xiaoming hdu 1002 A + B problem II, the conversion of numbers (inversion), the addition of large numbers ... ）