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[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("\n");    }    return 0;}