标签：ted RoCE esc sts cin line nat put sele
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
Author
Ignatius.L
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#include<iostream> #include<stdio.h> #include<string> using namespace std; int main() { string s1=""; string s2=""; string s3=""; int n; int length=0; cin>>n; int num=1; while(n--) { cin>>s1>>s2; string s4=s2; string s5=s1; if(s1.length()>s2.length()) { length=s1.length(); } else { s3=s1; s1=s2; s2=s3; length=s1.length(); } s3=string(s1.length()-s2.length(),‘0‘); s2=s3+s2; //cout<<s1<<s2; string temp="0"; int t=0; for(int i=length-1;i>=0;i--) { t=((s1[i]-48)+(s2[i]-48)+(temp[0]-48)); s1[i]=t%10+48; if(t>=10) temp[0]=‘1‘; else temp[0]=‘0‘; } cout<<"Case"<<" "<<num++<<":"<<endl; if(t>=10) cout<<s5<<" + "<<s4+" = "<<"1"<<s1<<endl; else cout<<s5<<" + "<<s4+" = "<<s1<<endl; if(n!=0) cout<<endl; } return 0; } //注意!!!!!!!!!!!!!!!!!!!!!!!!!!最后一行不要输出空行
标签：ted RoCE esc sts cin line nat put sele
原文地址：https://www.cnblogs.com/wzmm/p/12248726.html
