NYOJ 103 A+B Problem II
一个大数加法问题
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描述
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I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.
A,B must be positive.
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输入
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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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输出
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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.
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样例输入
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2 1 2 112233445566778899 998877665544332211
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样例输出
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Case 1: 1 + 2 = 3 Case 2: 112233445566778899 + 998877665544332211 = 1111111111111111110
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//代码如下
#include
#include
#include
using namespace std;
int main()
{
int n,s_m1[1010],s_m2[1010],s[1010],len1,len2,max,i,j,nCase=1;
string m1,m2;
cin>>n;
while(n--)
{
cin>>m1>>m2;
memset(s_m1,0,sizeof(s_m1));
memset(s_m2,0,sizeof(s_m2));
memset(s,0,sizeof(s));
len1=m1.length();
len2=m2.length();
for(i=0;ilen2) ? len1:len2;
j=0;
for(i=0;i0)
cout<