高精度板子

从网上当的

//__int128的这个板子要关同步流

#define int __int128
#define gc getchar()
#define cin(a) a=read()
int read(){
    int x=0;bool fl=0;char s=gc;
    while(!isdigit(s)){if(s=='-')fl=1;s=gc;}  // 处理符号位
    while(isdigit(s))x=(x<<1)+(x<<3)+s-'0',s=gc;  // 计算数值，相当于x = x*10 + (s-'0')
    return fl?-x:x;
}
void Print(int x){
//今天发现这个Print函数要加上下边x<0的情况，否则会输出神秘字符
    if(x<0){
        putchar('-');
        x=-x;
    }
    if(x/10) Print(x/10);  // 递归输出高位
    putchar(x%10+'0');     // 输出当前位
}

const int maxn = 50000;  
  
struct bign{  
    int d[maxn], len;  
  
    void clean() { while(len > 1 && !d[len-1]) len--; }  
  
    bign()          { memset(d, 0, sizeof(d)); len = 1; }  
    bign(int num)   { *this = num; }   
    bign(char* num) { *this = num; }  
    bign operator = (const char* num){  
        memset(d, 0, sizeof(d)); len = strlen(num);  
        for(int i = 0; i < len; i++) d[i] = num[len-1-i] - '0';  
        clean();  
        return *this;  
    }  
    bign operator = (int num){  
        char s[20]; sprintf(s, "%d", num);  
        *this = s;  
        return *this;  
    }  
  
    bign operator + (const bign& b){  
        bign c = *this; int i;  
        for (i = 0; i < b.len; i++){  
            c.d[i] += b.d[i];  
            if (c.d[i] > 9) c.d[i]%=10, c.d[i+1]++;  
        }  
        while (c.d[i] > 9) c.d[i++]%=10, c.d[i]++;  
        c.len = max(len, b.len);  
        if (c.d[i] && c.len <= i) c.len = i+1;  
        return c;  
    }  
    bign operator - (const bign& b){  
        bign c = *this; int i;  
        for (i = 0; i < b.len; i++){  
            c.d[i] -= b.d[i];  
            if (c.d[i] < 0) c.d[i]+=10, c.d[i+1]--;  
        }  
        while (c.d[i] < 0) c.d[i++]+=10, c.d[i]--;  
        c.clean();  
        return c;  
    }  
    bign operator * (const bign& b)const{  
        int i, j; bign c; c.len = len + b.len;   
        for(j = 0; j < b.len; j++) for(i = 0; i < len; i++)   
            c.d[i+j] += d[i] * b.d[j];  
        for(i = 0; i < c.len-1; i++)  
            c.d[i+1] += c.d[i]/10, c.d[i] %= 10;  
        c.clean();  
        return c;  
    }  
    bign operator / (const bign& b){  
        int i, j;  
        bign c = *this, a = 0;  
        for (i = len - 1; i >= 0; i--)  
        {  
            a = a*10 + d[i];  
            for (j = 0; j < 10; j++) if (a < b*(j+1)) break;  
            c.d[i] = j;  
            a = a - b*j;  
        }  
        c.clean();  
        return c;  
    }  
    bign operator % (const bign& b){  
        int i, j;  
        bign a = 0;  
        for (i = len - 1; i >= 0; i--)  
        {  
            a = a*10 + d[i];  
            for (j = 0; j < 10; j++) if (a < b*(j+1)) break;  
            a = a - b*j;  
        }  
        return a;  
    }  
    bign operator += (const bign& b){  
        *this = *this + b;  
        return *this;  
    }  
  
    bool operator <(const bign& b) const{  
        if(len != b.len) return len < b.len;  
        for(int i = len-1; i >= 0; i--)  
            if(d[i] != b.d[i]) return d[i] < b.d[i];  
        return false;  
    }  
    bool operator >(const bign& b) const{return b < *this;}  
    bool operator<=(const bign& b) const{return !(b < *this);}  
    bool operator>=(const bign& b) const{return !(*this < b);}  
    bool operator!=(const bign& b) const{return b < *this || *this < b;}  
    bool operator==(const bign& b) const{return !(b < *this) && !(b > *this);}  
  
    string str() const{  
        char s[maxn]={};  
        for(int i = 0; i < len; i++) s[len-1-i] = d[i]+'0';  
        return s;  
    }  
};  
  
istream& operator >> (istream& in, bign& x)  
{  
    string s;  
    in >> s;  
    x = s.c_str();  
    return in;  
}  
  
ostream& operator << (ostream& out, const bign& x)  
{  
    out << x.str();  
    return out;  
}