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
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1 #include<iostream> 2 #include<string> 3 using namespace std; 4 5 #define P(x) \ 6 cout << #x " " << x << ":" << endl; 7 8 #define PE(a, b, c, sum) \ 9 cout << a << " + " << b << " = " << (c ? "1" : "") << sum << endl; 10 int main() 11 { 12 int T; 13 cin >> T; 14 int Case = 1; 15 while(T--) 16 { 17 P(Case); 18 ++Case; 19 string a, b, sum; 20 cin >> a >> b; 21 //cout << a << " + " << b << " = " ; 22 int l_a = a.size()-1, l_b = b.size()-1, c = 0; 23 sum = (l_a > l_b ? a : b); 24 int i, j; 25 for(i = l_a, j = l_b; i >= 0 && j >= 0; --i, --j) 26 { 27 int s = a[i] - '0' + b[j] - '0' + c; 28 sum[i>j?i:j] = s % 10 + 48; 29 c = s / 10; 30 } 31 while(i >= 0) 32 {//处理进位 33 int s = sum[i] - '0' + c; 34 sum[i] = s % 10 + 48; 35 c = s / 10; 36 i--; 37 } 38 while(j >= 0) 39 {//处理 进位 40 int s = sum[j] - '0' + c; 41 sum[j] = s % 10 + 48; 42 c = s / 10; 43 j--; 44 } 45 // cout << (c ? "1" : "") << sum << endl; 46 PE(a, b, c, sum); 47 if(T) 48 cout << endl; 49 } 50 return 0; 51 }
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#include<iostream> #include<string> using namespace std; const int maxn = 500; #define P(x) \ cout << #x ": " << x << endl; int main() { string a; cin >> a; //string str; //string str("shi"); string str(maxn, 'a'); //str = ""; //长度也变为了0 for(int i = 0; i != a.size(); ++i) //the same above //for(int i = a.size()-1; i >= 0; --i) str[i] = a[i]; P(a); P(str); return 0; } /* str: "" output: "" str:"shi" output: san 由此可见,string 对象是有默认长度的(根据其赋值对象的长度决定) 与字符数组类似的,只不过不需明确指明,可根据赋值对象动态的适应并 改变它。字符赋值,其实质是在原有的基础上修改,是有长度限制的。 但是后面以后str = "";以后长度也都变为了0.由此可见,使用字符赋值的 关键是如何确定他的可能长度。 */