HDU 1002 A + B Problem II

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

题解：

由于本题的数据规模太大，所以用一般的数据类型，不能储存。需要用字符串来模拟.

#include<stdio.h>
#include<string.h>
int main()
{
char a[1000],b[1000],c[1001];
int i,j=1,p=0,n,n1,n2;
scanf("%d",&n);
while(n)
{
scanf("%s %s",a,b);
printf("Case %d:\n",j);
printf("%s + %s = ",a,b);
n1=strlen(a)-1;
n2=strlen(b)-1;
for(i=0;n1>=0||n2>=0;i++,n1--,n2--)
{
if(n1>=0&&n2>=0){c[i]=a[n1]+b[n2]-'0'+p;}
if(n1>=0&&n2<0){c[i]=a[n1]+p;}
if(n1<0&&n2>=0){c[i]=b[n2]+p;}
p=0;
if(c[i]>'9'){c[i]=c[i]-10;p=1;}
}
if(p==1){printf("%d",p);}
while(i--)
{printf("%c",c[i]);}
j++;
if(n!=1){printf("\n\n");}
else {printf("\n");}
n--;
}
}

当然如果你学的JAVA就可以直接大数过了，上面说的就可以当没看到。