A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 147889 Accepted Submission(s): 27893
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
由于处理的数比较大超过了long long 型，所以要变为数组处理比较好！
解题代码：
View Code
1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 using namespace std; 5 char a[10003], b[10003], c[10003]; 6 7 void deal(int lena, int lenb) 8 { 9 if (lena > lenb) 10 { 11 lenb --; 12 for (int i = lena - 1; i >= 0; i--) 13 { 14 //cout << b[lenb]<<endl; 15 if(lenb >= 0) 16 { 17 b[i] = b[lenb]; 18 b[lenb] = '0'; 19 lenb --; 20 } 21 else b[i] = '0'; 22 } 23 return; 24 } 25 if (lena < lenb) 26 { 27 lena--; 28 for (int i = lenb - 1; i >= 0; i --) 29 { 30 if (lena >= 0) 31 { 32 a[i] = a[lena]; 33 a[lena] = '0'; 34 lena --; 35 } 36 else a[i] = '0'; 37 } 38 return; 39 } 40 return ; 41 } 42 int main() 43 { 44 int n, num = 1; 45 int lena, lenb; 46 cin >> n; 47 while (n--) 48 { 49 cin >> a >> b; 50 lena = strlen (a); 51 lenb = strlen (b); 52 printf ("Case %d:\n", num++); 53 cout << a <<" + " << b<<" = "; 54 deal(lena, lenb); 55 int d = 0; 56 int j = 0; 57 for (int i = (lena > lenb ? lena : lenb) - 1; i >= 0; i--) 58 { 59 int temp = a[i] - 48 + b[i] - 48 + d; 60 d = temp/10; 61 c[j] = temp%10 + 48; 62 j++; 63 } 64 c[j] = d; 65 if (d) 66 { 67 c[j] = d + 48; 68 c[j+1] = 0; 69 } 70 for (int i =j-1; i >= 0; i --) 71 cout << c[i]; 72 cout << endl; 73 if (n) 74 cout <<endl; 75 } 76 return 0; 77 }