（大一题） 不知道错哪了

Question

题目是这样的：

时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 262144K，其他语言524288K
64bit IO Format: %lld
题目描述

 
This is a simple question.

Given you two hexadecimal digits x and y, you 
should determine whether 2x+10 is greater than 3y+5.
输入描述:
Each test contain multiple test cases. The first line contains 
the number of test cases t (1≤t≤100). Description of the test cases follow.

The description of each test case consists of two 
hexadecimal digits x, y separated by spaces.

It's guarantee that x ,y only consists of number '0'-'9' 
and capital letters 'A'-'F' and there are no leading zeros.

1≤x,y<16^1000

 输出描述:
For each test case, if 2x+10 > 3y+5, print "Yes",
 Otherwise print "No".

示例1
输入
3
A B
F0 E11
FF0123FFFFFFFFFF 01FEA23FFF
输出
No
No
Yes

我的代码是这样的：

#include<stdio.h>
#include<string.h>
int turn(char a, int t)
{
    if (a == 'A' || a == 'a') return 10 * t;
    else if (a == 'B' || a == 'b') return 11 * t;
    else if (a == 'C' || a == 'c') return 12 * t;
    else if (a == 'D' || a == 'd') return 13 * t;
    else if (a == 'E' || a == 'e') return 14 * t;
    else if (a == 'F' || a == 'f') return 15 * t;
    else return (a - '0') * t;
}
char retu(int t)
{
    if (t >= 0 && t <= 9) return t + '0';
    else if (t == 10) return 'A';
    else if (t == 11) return 'B';
    else if (t == 12) return 'C';
    else if (t == 13) return 'D';
    else if (t == 14) return 'E';
    else return 'F';
}
int main()
{
    int t;
    scanf("%d", &t);
    while (t--)
    {
        char a[1005] = { 0 }, b[1005] = { 0 };
        char ta[1005] = { 0 }, tb[1005] = { 0 };
        scanf("%s %s", a, b);
        int loa = strlen(a);
        int lob = strlen(b);
        int ia[1005] = { 0 }, ib[1005] = { 0 };
        for (int i = 0; i <= loa; i++)//处理x
        {
            if(i < loa)
                ia[i] += turn(a[i], 2);
            if (i == 0)
                ia[i] += 10;//转成运算后的数字
            while (ia[i] >= 16)//判断进位
            {
                ia[i] -= 16;
                ia[i + 1] += 1;
            }
            if(i < loa || i == loa && ia[i] != 0)//防止前导0出现影响长度
                ta[i] = retu(ia[i]);
        }
        for (int i = 0; i <= lob; i++)//处理y
        {
            if(i < lob)
                ib[i] += turn(b[i], 3);//同x
            if (i == 0)
                ib[i] += 5;
            while (ib[i] >= 16)
            {
                ib[i] -= 16;
                ib[i + 1] += 1;
            }
            if(i < lob || i == lob && ib[i] != 0)
                tb[i] = retu(ib[i]);
        }
        int lowa = strlen(ta), lowb = strlen(tb);
        if (lowa > lowb)//长度大的必然大
            printf("Yes\n");
        else if (lowa < lowb)
            printf("No\n");
        else
        {
            int u;
            for (u = lowa - 1; u >= 0; u--)//由于存的时候是从位置0开始存的，所以位置0是最小位，应从最后一位开始比
            {
                if (ta[u] > tb[u])
                {
                    printf("Yes\n");
                    break;
                }
                else if(ta[u] < tb[u])
                {
                    printf("No\n");
                    break;
                }
                else continue;
            }
            if (u == -1)//相等的情况
                printf("No\n");
        }
    }
}

没有通过所有测试样例

求解

Answer 1

char retu(int t)
{
    if (t >= 0 && t <= 9) return t + '0';
    else
        return 'A'+t-10;
}