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
思路分析:
共分三步进行处理,1,两数组各自从最高下标相加并存如新数组中,直到其中一个到达下标为0;
2,将下标没到达0的数组加到新数组中;
3,判断最后是否还有进位,若有加到新数组中。
#include<stdio.h>
#include<string.h>
int main()
{
int t;
int i,j,k,m,l1,l2;
char a[1005],b[1005];
int c[1005];
scanf ("%d",&t);
int x=1;
while (t--)
{
scanf ("%s%s",a,b);
int m = 0;
l1 = strlen(a);
l2 = strlen(b);
for (i=l1-1,j=l2-1,k=0; i>=0 && j>=0; i--,j--)
{
m = a[i]-48+b[j]-48+m;
c[k++] = m%10;
m /= 10;
}
if (k == l1)
{
while (j>=0)
{
m = b[j--]-48+m;
c[k++] = m%10;
m = m/10;
}
}
else
{
while (i>=0)
{
m = a[i--]-48+m;
c[k++] = m%10;
m = m/10;
}
}
if (m != 0)
{
c[k++] = m;
}
printf ("Case %d:\n",x++);
printf ("%s + %s = ",a,b);
for (i=k-1; i>=0; i--)
{
printf ("%d",c[i]);
}
printf ("\n");
if (t)
printf ("\n");
}
return 0;
}