hdu1002 A + B Problem II(大数加法一)模板
A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 111273 Accepted Submission(s): 21079
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
2
1 2
112233445566778899 998877665544332211
Sample Output
Case 1:
1 + 2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110
分析:使用string 可以返回大数 大数加法模板
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<iostream> 2 #include<string> 3 #include<fstream> 4 #define N 1000 5 using namespace std; 6 7 8 string bigmult(string a,string b) 9 { 10 int la,lb,lc,i,j,flag; 11 string c=""; 12 char t; 13 i=la=a.length()-1; 14 j=lb=b.length()-1; 15 flag=0; 16 while(i>=0&&j>=0) 17 { 18 t=a[i]+b[j]+flag-'0'; 19 if(t>'9') 20 { 21 t=t-10;flag=1; 22 } 23 else flag=0; 24 c+=t; 25 i--;j--; 26 } 27 while(i>=0) 28 { 29 t=a[i]+flag; 30 if(t>'9') 31 { 32 t=t-10;flag=1; 33 } 34 else flag=0; 35 c+=t; 36 i--; 37 } 38 while(j>=0) 39 { 40 t=b[j]+flag; 41 if(t>'9') 42 { 43 t=t-10;flag=1; 44 } 45 else flag=0; 46 c+=t; 47 j--; 48 } 49 if(flag) c+=(flag+'0'); 50 lc=c.length(); 51 for(i=0,j=lc-1;i<j;i++,j--) 52 { 53 t=c[i];c[i]=c[j];c[j]=t; 54 } 55 return c; 56 } 57 58 string ans[N]; 59 60 int main() 61 { 62 int test,k=1; 63 string a,b; 64 cin>>test; 65 while(test--) 66 { 67 cin>>a>>b; 68 cout<<"Case "<<k++<<":"<<endl; 69 cout<<a<<" + "<<b<<" = "<<bigmult(a,b)<<endl; 70 if(test)cout<<endl; 71 } 72 return 0; 73 }