HDu 2100 Lovekey(進数での高精度加算)

タイトル接続:2100 Lovekey
問題の構想を解く:直接加算シミュレーション、ただ10を26に変えます.

#include <stdio.h>
#include <string.h>
const int N = 205;

int change(char *str, int num[]) {
    int len = strlen(str);
    memset(num, 0, sizeof(num));
    for (int i = 0; i < len; i++)
	num[len - i - 1] = str[i] - 'A';
    return len;
}

int add(int a[], int b[], int sum[], int na, int nb, int base) {
    int n = 0, t = 0;
    memset(sum, 0, sizeof(sum));
    for (int i = 0; i < na || i < nb; i++) {
	sum[i] = t;
	if (i < na) sum[i] += a[i];
	if (i < nb) sum[i] += b[i];
	t = sum[i] / base;
	sum[i] %= base;
	n++;
    }

    while (t) {
	sum[n++] = t % base;
	t = t / base;
    }
    return n;
}

int main() {
    int n1, n2, num1[N], num2[N], sum[N];
    char str1[N], str2[N];
    while (scanf("%s%s", str1, str2) == 2) {
	n1 = change(str1, num1);
	n2 = change(str2, num2);

	n1 = add(num1, num2, sum, n1, n2, 26);

	int flag = 0;
	for (int i = n1 - 1; i >= 0; i--) {
	    if (flag || sum[i]) {
		printf("%c", 'A' + sum[i]);
		flag = 1;
	    }
	}
	if (flag == 0)	printf("A");
	printf("
");
    }
    return 0;
}