ＨＤＵ　A + B Problem IIA + 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

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

Sample Output

字元串模拟大數加法

#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;

}

本文轉自NewPanderKing51CTO部落格，原文連結：http://www.cnblogs.com/newpanderking/archive/2011/07/31/2122519.html ，如需轉載請自行聯系原作者

ＨＤＵ　A + B Problem IIA + B Problem II

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ＨＤＵ A + B Problem IIA + B Problem II

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ＨＤＵ　A + B Problem IIA + B Problem II