（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 return 'A'+t-10;
}
int main()
{
    int t;
    scanf("%d", &t);
    getchar();
    while (t--)
    {
        char a[1005] = { 0 }, b[1005] = { 0 };
        char ta[1005] = { 0 }, tb[1005] = { 0 };
        char ch;
        int answer=0;
        for(int g=0;;)//处理前导零
        {
            scanf("%c",&ch);
            if(ch==' ') break;
            else
            {
                if(answer==0)
                {
                    if(ch=='0') continue;
                    else answer=1;
                }
                if(answer==1)
                {
                    a[g]=ch;
                    g++;
                }
            }
        }
        answer=0;
        for(int g=0;;)//处理前导零
        {
            scanf("%c",&ch);
            if(ch=='\n') break;
            else
            {
                if(answer==0)
                {
                    if(ch=='0') continue;
                    else answer=1;
                }
                if(answer==1)
                {
                    b[g]=ch;
                    g++;
                }
            }
        }
        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)//防止添加前导0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");
        }
    }
}

报错是这样的.:

求解

Answer 1

#include<stdio.h>
#include<string.h>
int turn(char a, int t) {
    if (a >= '0' && a <= '9') return (a - '0') * t;
    else return (a - 'A' + 10) * t ;
}
char retu(int t) {
    if (t >= 0 && t <= 9) return t + '0';
    else return 'A' + t - 10;
}
char a[1005] = { 0 }, b[1005] = { 0 };
//char ta[1005] = { 0 }, tb[1005] = { 0 };
int ia[1005] = { 0 }, ib[1005] = { 0 };
int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        memset(a, 0, sizeof(a));
        memset(b, 0, sizeof(b));
        memset(ia, 0, sizeof(ia));
        memset(ib, 0, sizeof(ib));
        scanf("%s %s", a, b);
        int loa = strlen(a);
        int lob = strlen(b);
        int lia = 0, lib = 0;
        for (int i = loa-1; i >=0; i--) { //处理x
            ia[lia] += turn(a[i], 2);
            if (i == loa - 1)
                ia[lia] += 5;//转成运算后的数字
            ia[lia + 1] += ia[lia] / 16;
            ia[lia] %= 16;
            lia++;
        }
        if(ia[lia]>0){
            lia++;
        }
        while(ia[lia] >= 16) {
            ia[lia + 1] += ia[lia] / 16;
            ia[lia] %= 16;
            lia++;
        }
        while(ia[lia]==0) lia--;
        for (int i = lob-1; i >=0; i--) { //处理y
            ib[lib] += turn(b[i], 3);
            ib[lib + 1] += ib[lib] / 16;
            ib[lib] %= 16;
            lib++;
        }
        if(ia[lib]>0){
            lib++;
        }
        while(ib[lib] >= 16) {
            ib[lib + 1] += ib[lib] / 16;
            ib[lib] %= 16;
            lib++;
        }
        while(ib[lib]==0) lib--;
//        int lowa = strlen(ta), lowb = strlen(tb);
        if (lia > lib)//长度大的必然大
            printf("Yes\n");
        else if (lia < lib)
            printf("No\n");
        else {
            int u;
            for (u = lia - 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 (ia[u] > ib[u]) {
                    printf("Yes\n");
                    break;
                } else if(ia[u] < ib[u]) {
                    printf("No\n");
                    break;
                } else continue;
            }
            if (u == -1)//相等的情况
                printf("No\n");
        }
    }
    return 0;
}