A + B Problem II |
| Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) |
| Total Submission(s): 307 Accepted Submission(s): 147 |
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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.
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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.
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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. |
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Sample Input
2
1 2
112233445566778899 998877665544332211
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Sample Output
Case 1:
1 + 2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110
字符串模拟大数加法
#include <iostream>
#include <string.h> #include <stdio.h> using namespace std; char sum[1000]; void add(string a,string b) { string c ; if(a.length()<b.length()) {c = a ; a = b; b = c;} int l1 = a.length()-1; int l2 = b.length()-1; int k=0,i=0; memset(sum,0,sizeof(sum)); char t; while(l2!=-1) { t = (a[l1]-'0')+(b[l2]-'0'); sum[i]= (t+k)%10+'0'; k = (t+k)/10; l1--; l2--; i++; } while(l1!=-1) { t = (a[l1]-'0'); sum[i] = (k+t)%10+'0'; k = (t+k)/10; l1--; i++; } if(k!=0)sum[i]=k+'0'; } int main() { string a,b; int T; cin>>T; for(int i = 1 ; i <= T ;i ++) { cin>>a>>b; add(a,b); cout<<"Case "<<i<<":"<<endl; cout<<a<<" + "<<b<<" = "; for(int j = strlen(sum)-1; j >=0;j--) { cout<<sum[j]; } if(i<T) cout<<endl<<endl; else cout<<endl; } return 0; }
本文转自NewPanderKing51CTO博客，原文链接：
http://www.cnblogs.com/newpanderking/archive/2011/07/31/2122519.html
，如需转载请自行联系原作者
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ＨＤＵ A + B Problem II
2017-11-15
1207
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