A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 368505 Accepted Submission(s): 71806
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
解毒了终于,我好累,,
#include <iostream> #include<string> using namespace std; int main(int argc, char** argv) { string str1,str2; int t;cin>>t; int cont=0; while(t--) { int p; int a[1005]={0}; if(cont>0) cout<<endl; cont++; cout<<"Case"<<" "<<cont<<":"<<endl; cin>>str1>>str2; int len1=str1.size() ; int len2=str2.size() ; int i,j,k; j=len2-1; for( i=len1-1,k=0;i>=0;i--) { p=str1[i]-2*'0'+str2[j]; if(p+a[k]>=10) { a[k]+=(p-10); a[k+1]+=1; } else a[k]+=p; k++; j--; if(j<0) break; } int g; if(i<j) { cout<<str1<<" "<<"+"<<" "<<str2<<" "<<"="<<" "; for(int h=len2-len1-1;h>=0;h--) { if(a[k]+(str2[h]-'0')>=10) { a[k]+=(str2[h]-'0'); a[k]-=10; a[k+1]++; } else a[k]+=(str2[h]-'0'); k++; } for(g=k;g>=0;g--) { if(a[g]!=0) break; } if(g==-1) cout<<"0"; else for(int i=g;i>=0;i--) cout<<a[i]; cout<<endl; } else { cout<<str1<<" "<<"+"<<" "<<str2<<" "<<"="<<" "; for(int h=len1-len2-1;h>=0;h--) { if(a[k]+(str1[h]-'0')>=10) { a[k]+=(str1[h]-'0'); a[k]-=10; a[k+1]++; } else a[k]+=(str1[h]-'0'); k++; } for(g=k;g>=0;g--) { if(a[g]!=0) break; } if(g==-1) cout<<"0"; else for(int i=g;i>=0;i--) cout<<a[i]; cout<<endl; } } return 0; }