hdu 2100 Lovekey(进制下的高精度加法）

题目连接：2100 Lovekey

解题思路：直接加法模拟， 只是将10 换成26.

#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
, num2
, sum
;
char str1
, str2
;
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("\n");
}
return 0;
}