HDU__1002A + B Problem II(大数加法)
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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
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
注意事项都在代码里注释了,还是爱犯很多小毛病
#include<iostream> #include<string> #include<cstdio> #include<vector> #include<algorithm> using namespace std; int main() { int t,putcnt = 0; string s1,s2; vector<int> a,b,c; cin>>t; while(t--) { cin>>s1>>s2;// a.clear();// b.clear();// c.clear();//清空后并没有自动赋值为0!! //逆序 ——并没有实现逆序!!! for(int i = s1.size() - 1; i >= 0;i--) { a.push_back(s1[i] - '0');}//for(int i = 0; i < a.size();i++)//{//cout<<a[i];// } for(int i = s2.size() - 1; i >= 0;i--) { b.push_back(s2[i] - '0');}//cout<<endl;//for(int i = 0; i < b.size();i++)//{//cout<<b[i];// } //相加到s3 int is_jin = 0,push;int lenmax = max(a.size(),b.size());for(int i = 0;i < lenmax;i++){if(i >= a.size())//补位操作 {a.push_back(0);}if(i >= b.size()){b.push_back(0);}push = (a[i] + b[i] + is_jin) % 10;c.push_back(push);is_jin = (a[i] + b[i] + is_jin) / 10;}if(is_jin){c.push_back(1);}// 输出printf("Case %d:\n",++putcnt);cout<<s1<<" + "<<s2<<" = ";//string只支持cin和cout不能用scanf和printf for(int i = c.size() - 1;i >= 0;i--){cout<<c[i];} cout<<"\n";//注意输出格式 if(t >= 1) { cout<<endl; } //尾部虽然都要求空一个空行,但是不能空两个啊!!~~注意输出格式,好好理解题目吧 a.clear(); b.clear(); c.clear(); s1.clear(); s2.clear(); } return 0; }
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