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高精度模板.

#include <iostream>
#include <string>
#include <cstring>
#include <cstdio>
using namespace std;

const int maxn = 1000;

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; } int main() { freopen("noip.in","r",stdin); freopen("noip.out","w",stdout); bign s=0,t; while (cin>>t) { if (t.len==1&&!t.d[0]) break; s=s+t; } cout<<s<<endl; return 0; }

 寫了個沒有/和後面運算的簡單一點的。。都是自己可以理解語法

/也挺簡單 但感覺noip用不到

#include <bits/stdc++.h>
using namespace std;
const int maxn=1000;
#define IL inline
struct bign{
    int d[maxn],len;
    IL void clear() {while (len>1&&!d[len-1]) len--;}
    bign() {len=1;}
    bign(int num)     { *this = num; } 
    bign(char* num) { *this = num; }
    bign operator = (const char *num)
    {
        len=strlen(num);
        for (int i=0;i<len;i++) d[i]=num[len-i-1]-'0';
        clear();
        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],c.d[i+1]+=c.d[i]/10,c.d[i]%=10;
        c.len=max(c.len,b.len);
        if (c.d[c.len]) c.len++;
        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+1]--,i++;
        return c;
    }
    bign operator *(const bign b)
    {
        int i,j; bign c; c.len=b.len+len;
        for (j=0;j<b.len;j++)
          for (i=0;i<len;i++)
            c.d[i+j]+=b.d[j]*d[i];
        for (i=0;i<c.len;i++)
          c.d[i+1]+=c.d[i]/10,c.d[i]%=10;
        c.clear();
        return c;
    }
    string str()
    {
        char s[maxn]={};
        for (int i=0;i<len;i++) s[len-i-1]=d[i]+'0';
        return s;
    }
};
int main()
{
    bign a=3424,b=213123,c=324,d=123,e=1344,f=132;
    bign ans=(a+b)*(c+d)*(e-f);
    cout<<ans.str();
    return 0;
}