杭电_ACM_A + B Problem II

2
1 2
112233445566778899 998877665544332211

Case 1:
1 + 2 = 3

Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110

#include <stdio.h>
#include <string>
#define MAX 1005
int main()
{
int T, a[MAX], b[MAX], c[MAX], i, j, k, carry;
char sa[MAX], sb[MAX];
scanf("%d", &T);
for (j = 1; j <= T; j++)
{
scanf("%s %s", sa, sb);
memset(a, 0, sizeof(a));
memset(b, 0, sizeof(b));
carry = 0;
//a[0] is the length of a, it's more convenient
a[0] = strlen(sa);
b[0] = strlen(sb);
//translate the char to number and inverse
for (i = 1; i <= a[0]; i++)
{
a[i] = sa[a[0] - i] - '0';
}
for (i = 1; i <= b[0]; i++)
{
b[i] = sb[b[0] - i] - '0';
}
//get the longer one for result
c[0] = a[0] > b[0] ? a[0] : b[0];
//add one by one and note the carry
for (i = 1; i <= c[0]; i++)
{
c[i] = (a[i] + b[i] + carry) % 10;
carry = (a[i] + b[i] + carry) / 10;
}
//if the last carry is not zero, you must add the one.
if (carry != 0)
{
c[0]++;
c[c[0]] = carry;
}
//formatted printing
printf("Case %d:\n", j);
for (k = a[0]; k >= 1; k--)
printf("%d", a[k]);
printf(" + ");
for (k = b[0]; k >= 1; k--)
printf("%d", b[k]);
printf(" = ");
for (k = c[0]; k >= 1; k--)
printf("%d", c[k]);
printf("\n");
if (j != T)
printf("\n");
}
return 0;
}