csu1002 A+B(III)

题目描述：

There are multiple test cases. Each test case contains only one line. Each line consists of a pair of integers a and b（1=< a,b <=10¹⁶）, separated by a space. Input is followed by a single line with a = 0, b = 0, which should not be processed.

For each pair of input integers a and b you should output the sum of a and b in one line, and with one line of output for each line in input.

RE : 数组开的不够大

看到有的代码只有114B（我的超1000B），不解。

// csuoj 1002 : A+B(III)
# include <stdio.h>
# include <string.h>
# define MAXN 20                // RE: # define MAXN 17
char a[MAXN], b[MAXN], ans[MAXN];
void add(char *a, char *b, char *ans); //ans=a+b
void strnrev(char *a);
int main()
{
    while(2==scanf("%s%s", a,b))
    {
        if(1==strlen(a) && 1==strlen(b) && a[0]=='0' && b[0]=='0') return ;
        strnrev(a);
        strnrev(b);
        add(a, b, ans);
        strnrev(ans);
        printf("%s\n", ans);
    }
    return 0;
}
void add(char *a, char *b, char *ans)
{
    char *p, *q;
    int i, j, tmp, c;
    if(strlen(a)>strlen(b)) {p=a;q=b;}
    else {p=b;q=a;}
    i = c = 0;
    while (q[i] != '\0')
    {
        tmp = (q[i]-'0')+(p[i]-'0')+c;
        ans[i] = tmp%10+'0';
        c = tmp/10;
        ++i;
    }
    while (p[i] != '\0')
    {
        tmp = (p[i]-'0')+c;
        ans[i] = tmp%10+'0';
        c = tmp/10;
        ++i;
    }
    if(c==1) ans[i++] = c+'0';
    ans[i] = '\0';
}
void strnrev(char *a)
{
    int i, len;
    char tmp;
    len = strlen(a);
    for (i = 0;i < len>>1; ++i)
    {
        tmp = a[i];
        a[i] = a[len-i-1];
        a[len-i-1] = tmp;
    }
}