hdoj1002A + B Problem II（大数相加）

[align=left]Problem Description[/align]
I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.

 

[align=left]Input[/align]
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.

 

[align=left]Output[/align]
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.

 

[align=left]Sample Input[/align]

2
1 2
112233445566778899 998877665544332211

 

[align=left]Sample Output[/align]

Case 1:
1 + 2 = 3

Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110

 

[align=left]Author[/align]
Ignatius.L
 
代码：

#include<stdio.h>
#include<string.h>
int main()
{
int n;
scanf("%d",&n);
int k=1;
while(n--)
{
char a[1001],b[1001],c[1001];
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
memset(c,0,sizeof(c));
scanf("%s %s",a,b);
int s1=strlen(a),s2=strlen(b),p=0;
if(s1<s2)
{
int t;
t=s1;
s1=s2;
s2=t;
p++;
}
for(int i=s1-1,j=s2-1;i>=0;i--,j--)
{
if(p==0)
{
if(j>=0)
{
c[i+1]+=(a[i]+b[j]-'0'-10);
if(c[i+1]<48)
{
c[i+1]+=10;
}
else
c[i]++;
}
else
{
c[i+1]+=(a[i]-10);
if(c[i+1]<48)
{
c[i+1]+=10;
}
else
c[i]++;
}
}
else
{
if(j>=0)
{
c[i+1]+=a[j]+b[i]-'0'-10;
if(c[i+1]<48)
{
c[i+1]+=10;
}
else
c[i]++;
}
else
{
c[i+1]+=b[i]-10;
if(c[i+1]<48)
{
c[i+1]+=10;
}
else
c[i]++;
}
}
}
printf("Case %d:\n%s + %s = ",k++,a,b);
if(c[0]!=0)
printf("%d",c[0]);
for(int i=1;i<=s1;i++)
printf("%c",c[i]);
if(n!=0)
printf("\n\n");
else
printf("\n");
}
return 0;
}

思路：这道题用两个数组保存大数，然后用另一个数组保存每个位置相加的数。超过十就-10，前一位加一。最后输出注意格式还有过程中注意最后一位是否大于十，大于的话还要进一位。