算法训练 回文数

#include <stdio.h> 
#include <string.h>
#include <stdlib.h>
#define M 30
char str[M],str1[M];
int len;
int Ispalindrome()//判断回文数
{
    int i;
    for(i=0;i<len/2;i++)
        if(str[i] != str[len-i-1])
            return 0;
    return 1;
}
int main()
{
    int n,i,j,t;
    int t1,flag;
    char temp[2] = {‘\0‘};
    scanf("%d%s",&n,&str);
    len = strlen(str);
    j = 0;
    flag = 0;if(n == 16)
    {
        for(t=0;t<30;t++)//30步以内得到回文数 
        {
            if(!Ispalindrome())
            {
                for(i=0;i<len;i++)
                {    
                    t1 = 0;
                    if(str[i]>=‘0‘ && str[i] <=‘9‘)
                        t1 += (str[i] - ‘0‘);
                    else
                        t1 += (str[i] - ‘A‘ + 10);
                    if(str[len-i-1]>=‘0‘ && str[len-i-1] <=‘9‘)
                        t1 += (str[len-i-1] - ‘0‘);
                    else
                        t1 += (str[len-i-1] - ‘A‘ + 10);
                    t1 += flag;
                    if(t1 >= n)//要进位 
                    {
                        if(t1-n<= 9)
                            itoa(t1-n,temp,10);//整型转化为字符型
                        else
                            temp[0] = ‘A‘ + (t1-n-10);
                        flag = 1;
                        str1[j++] = temp[0];
                    }
                    else{
                        if(t1<= 9)
                            itoa(t1,temp,10);//整型转化为字符型
                        else
                            temp[0] = ‘A‘ + (t1-10);
                        flag = 0;
                        str1[j++] = temp[0];
                    }    
                }
                if(flag)
                {
                    str1[j] = ‘1‘;
                    str1[j+1] = ‘\0‘;
                    flag = 0;//置0
                }
                else
                    str1[j] = ‘\0‘;
                strcpy(str,str1);//将str1赋值给str
                len = strlen(str) ;
                j = 0;
            }
            else{
                printf("STEP=%d\n",t);
                break;
            }
        }
        if(t == 30)
            printf("Impossible!\n");
    }
    else{
        for(t=0;t<30;t++)//30步以内得到回文数 
        {
            if(!Ispalindrome())
            {
                for(i=0;i<len;i++)
                {
                    t1 = (str[i]-‘0‘) + (str[len-i-1] - ‘0‘) + flag;//要加flag 
                    if(t1 >= n)//要进位 
                    {
                        itoa(t1-n,temp,10);//整型转化为字符型
                        flag = 1;
                        str1[j++] = temp[0];
                    }
                    else{
                        itoa(t1,temp,10);//整型转化为字符型
                        flag = 0;
                        str1[j++] = temp[0];
                    }
                }
                if(flag)
                {
                    str1[j] = ‘1‘;
                    str1[j+1] = ‘\0‘;
                    flag = 0;//置0
                }
                else
                    str1[j] = ‘\0‘;
                strcpy(str,str1);//将str1赋值给str
                j = 0;
                len = strlen(str) ;
            }
            else{
                printf("STEP=%d\n",t);
                break;
            }
        }
        if(t == 30)
            printf("Impossible!\n");
    
    }
    return 0;
}