nyoj 103A+B Problem II

A+B Problem II
时间限制：3000 ms  |  内存限制：65535 KB
难度：3
描述
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.


输入
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.
输出
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.
样例输入
2
1 2
112233445566778899 998877665544332211
样例输出
Case 1:
1 + 2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110
来源
经典题目
上传者
张云聪

代码：

 #include<cstdio>#include<cstring>int main(){int t;scanf("%d",&t);for (int ca=1;ca<=t;ca++){char a[1020],b[1020];int aa[1020]={0},bb[1020]={0};scanf("%s%s",a,b);printf("Case %d:\n%s + %s = ",ca,a,b);int al=strlen(a);int bl=strlen(b);for (int i=0;i<al;i++)aa[i]=a[al-1-i]-'0';for (int i=0;i<bl;i++)bb[i]=b[bl-1-i]-'0';if (al>=bl)    for  (int ii=0;ii<al;ii++)    {    aa[ii]+=bb[ii];    if (aa[ii]>9&&ii!=al-1)    {    aa[ii+1]++;    aa[ii]%=10;    }    }    else    for (int ii=0;ii<bl;ii++)    {    bb[ii]+=aa[ii];    if (bb[ii]>9&&ii!=bl-1)    {    bb[ii+1]++;    bb[ii]%=10;    }    }    if (al>=bl)    for (int ii=al-1;ii>=0;ii--)    printf("%d",aa[ii]);    else    for (int ii=bl-1;ii>=0;ii--)    printf("%d",bb[ii]);    printf("\n");    } return 0;}