【回文数(二)】 （C语言代码）

解题思路:





注意事项:





参考代码:

#include<stdio.h>
#include<string.h>
int f(char *str);         //判断是否为回文数
int main()
{
    int N,i,j,nHighestPos,count,tem;
    char s[100]={0};             //存放M
    int a[100]={0},b[100]={0};     //计算M与M的逆序的和，存放在a中
    while(scanf("%d",&N)!=EOF)
    {
        memset(s,0,sizeof(s));    //初始化数组
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        count=0;
        getchar();
        gets(s);
        if(f(s))
        {
            printf("STEP=%d\n",0);
            for(i=0;i<strlen(s);i++)
                s[i]=0;
        }
        else
        {
            while(!f(s)&&count<30)
            {
                for(i=0,j=strlen(s)-1;i<strlen(s);i++)
                {
                    a[j--]=s[i]-'0';
                    b[i]=s[i]-'0';
                }
                for(i=0;i<strlen(s)+10;i++)
                {
                    a[i]+=b[i];
                    if(a[i]>=N)
                    {
                        a[i+1]+=1;
                        a[i]-=N;
                    }
                    if(a[i]) nHighestPos=i;
                }
                for(i=nHighestPos,j=0;i>=0;i--)
                    s[j++]=a[i]+'0';
                count++;
            }
            if(count<30) printf("STEP=%d\n",count);
            else printf("Impossible!\n");
        }
    }
    return 0;
}
int f(char *str)
{
    int len,i,j;
    char c[100]={0};
    len=strlen(str);
    for(i=len-1,j=0;i>=0;i--)
        c[j++]=str[i];
    for(i=0;i<len;i++)
        if(str[i]!=c[i]) break;
    if(i==len) return 1;
    else return 0;
}