海明码检验和纠错

/*   海明码检验与纠错.*/#include <cstdio>#include <cstring>#include <algorithm>using namespace std;/*    测试样例    110010100000    0011101    1110110    011001011001    0001001*/char s[105];  // 用来接收信息位字符串char ans[1005]; //得出的海明码存储在这里const int hehe[] = {1,2,4,8,16,32,64,128,256,512,1024}; //这个办法真是呵呵了。。。。const int hehe_len = sizeof(hehe) / sizeof(int); //呵呵数组的长度，懒得算int r_len;        //校验位的位数int k; //信息位个数int step; //在海明码中信息位的最高下标。int s1[105];int s1_len;struct Node{    int a[10];  //存储ri是由谁生成的（存下标）    int len ;  //a数组的长度    int val;  //校验位的值} r[105];int pow(int b,int r_len) //幂运算{    int ret = 1;    for(int i = 1; i <= r_len; ++i)        ret *= b;    return ret;}bool findPos(int n)  //找到hehe中的数{    for(int i = 0; i < hehe_len; ++i)        if(hehe[i] == n)            return true;    return false;}int f(char c)  //char -> int{    return c - '0';}int zh(){    int ret = 0;    for(int i = 0;i < r_len;++i)        if(s1[i])           ret += pow(2,i);    return ret;}void solve(){    int len_n = strlen(s + 1);    for(k = len_n;k >= 0;--k)        if(pow(2,len_n - k) >= len_n + 1)           {         r_len = len_n - k;         break;        }    memset(ans,-1,sizeof ans);    memset(r,0,sizeof r);    for(int i = 1,j = len_n; i <= j; ++i,--j) //逆序字符串,以便对应I1,I2....    {        char t = s[i];        s[i] = s[j];        s[j] = t;    }    //printf("%s\n",s);    r_len = 0;    for(int i = 1;i <= len_n;++i)    {        if(findPos(i))        {            r[r_len++].val = f(s[i]);        }    }    int i;    for(step = 3,i = 1;i <= len_n;)    {        if(!findPos(step))        {            if(findPos(i))++i;            else ans[step++] = s[i++];        }        else step++;    }    //printf("step = %d\n",step);    for(int j = step - 1; j >= 3; --j)   // 找出ri是由哪几位在一起生成的.    {        if(!findPos(j))        {            int t = j;            for(int l = hehe_len - 1; l >= 0 && t > 0; --l)            {                if(t >= hehe[l])                {                    t -= hehe[l];                    r[l].a[r[l].len++] = j;                }            }        }    }    for(int i = 1,j = 0;i <= max(step - 1,pow(2,r_len - 1));++i)    {        if(findPos(i))        {            ans[i] = r[j++].val + '0';        }    }    s1_len = 0;    for(int i = 0;i < r_len;++i)    {        s1[i] = r[i].val;        for(int j = 0;j < r[i].len;++j)        {            s1[i] = s1[i] ^ f(ans[r[i].a[j]]);        }    }    int pos = zh();    if(pos == 0)    {        printf("这个海明码没有错误！\n");    }    else    {        ans[pos] == '0' ? ans[pos] = '1' : ans[pos] = '0';        printf("第%d位出现错误,改正后如下\n",pos);        for(int i = max(step - 1,pow(2,r_len - 1));i >= 1 ;--i)            printf("%c",ans[i]);        printf("\n");    }}int main(){    while(~scanf("%s",s + 1)) solve();    return 0;}