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HDOJ1002-A + B Problem II(高精加)
Problem Description
I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.
Input
The 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 two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000.
Output
For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line is the an equation "A + B = Sum", Sum means the result of A + B. Note there are some spaces int the equation. Output a blank line between two test cases.
Sample Input
21 2112233445566778899 998877665544332211
Sample Output
Case 1:1 + 2 = 3?Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110
Means:
求大整数的A+B
Solve:
高精模板题
Code:
1 //仅当两个都为正 2 #include <iostream> 3 #include <cstdio> 4 #include <cstring> 5 using namespace std; 6 7 struct big_number 8 { 9 int digit[10000];10 int len;11 big_number()12 {13 memset(digit,0,sizeof(digit));14 len=0;15 }16 };17 18 big_number reset_number(char *in)//将低位存在数组前面，如123存到数组里就是32119 {20 big_number after;21 after.len=strlen(in);22 for(int i=0;i<after.len;++i)23 {24 after.digit[i]=(in[after.len-i-1]-48);25 }26 return after;27 }28 29 big_number add(big_number num1,big_number num2)30 {31 int carry=0;32 big_number ans;33 for(int i=0;i<num1.len || i<num2.len;++i)34 {35 int temp=num1.digit[i]+num2.digit[i]+carry;36 ans.digit[ans.len++]=temp%10;37 carry=temp/10;38 }39 if(carry!=0)40 ans.digit[ans.len++]=carry;41 return ans;42 }43 44 int main()45 {46 47 int t;48 scanf("%d" , &t);49 for(int c = 1 ; c <= t ; ++c)50 {51 //printf("%s\n%s\n",in1,in2);52 char in1[10000]={‘\0‘},in2[10000]={‘\0‘};53 scanf("%s%s",in1,in2);54 big_number change1=reset_number(in1);55 big_number change2=reset_number(in2);56 big_number ans=add(change1,change2);57 printf("Case %d:\n" , c);58 printf("%s + %s = " , in1 , in2);59 for(int i=ans.len-1;i>=0;--i)60 {61 printf("%d",ans.digit[i]);62 }63 if(c != t)64 printf("\n\n");65 else66 printf("\n");67 }68 return 0;69 }
HDOJ1002-A + B Problem II(高精加)