[D] hex PAT B1022 A + B

[D] hex PAT B1022 A + B

Enter two non-negative decimal integers A and B (<= 2 ^ 30-1) , the output A + D B (1 <D <= 10) Number of decimal.
Input format:
Input three integers given sequentially in a row A, B and D.
Output Format:
Output D-nary number of A + B.
Sample input:
1,234,568
Output Sample:
1103

#include <stdio.h>

//十进制转换为D进制（除基取余法）
int main() {
    int choose;
    printf("输入1 十进制转换为D进制\n输入2 D进制转换为十进制\n");
    scanf("%d", &choose);
    if (choose == 1) {
        int a, d;
        printf("输入十进制数 和将要转换的进制D\n");
        scanf("%d%d", &a, &d);
        int sum = a;
        int ans[31], num = 0;
        //取余数
        do {
            ans[num++] = sum % d;
            sum = sum / d;
        } while (sum != 0);
        //逆序输出
        for (int i = num - 1; i >= 0; i--) {
            printf("%d", ans[i]);
        }
    } else if (choose == 2) {
        int a, d;
        printf("输入D进制数 和进制D\n");
        scanf("%d%d", &a, &d);
        //将D进制的a转换为十进制的y
        int y = 0, product = 1;//表示D的0次方
        while (a != 0) {
            //逐次取出个位
            y = y + (a % 10) * product;
            //去掉已经计算的个位数
            a = a / 10;
            //修改权重 d^0 d^1...
            product = product * d;
        }
        printf("%d", y);
    }

    return 0;
}

Test Results: