高精度模板 洛谷Luogu P1932 A+B & A-B & A*B & A/B Problem

P1932 A+B & A-B & A*B & A/B Problem

题目背景

这个题目很新颖吧！！！

题目描述

求A、B的和差积商余！

输入输出格式

输入格式：

两个数两行

A B

输出格式：

五个数

和 差 积 商 余

输入输出样例

输入样例#1：

1

1

输出样例#1：

2

0

1

1

0

说明

length(A),length(B)<=10^4

每个点3s。

题目链接

很明显，这道题是一道模板题，是很明显的高精度算法。当我翻阅《算法竞赛入门经典（第二版）》时，帅气的汝佳哥告诉我在代码仓库中有减法、乘法、除法的代码，但是当我来到代码仓库时，却被眼前的景象惊呆了

并没有什么减乘除！！

那这简直是一把鼻涕一把眼泪啊！还要自己思考、自己想算法。没办法，于是打了一份模板（调试了好久啊喂(╯‵□′)╯︵┻━┻）。但可怕的是，发生了如下情况：



七十分代码：[真正的模板]

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string>
#include<vector>
#include<cmath>
#define ll long long
#define INF 2147483647
#define ll_INF 9223372036854775807
using namespace std;
struct BigInteger {
    static const int BASE = 10000;
    static const int WIDTH = 4;
    int size, s[11000];
    inline void killzero() {
        while( s[size - 1] == 0 && size > 1 ) size --;
    }
    inline void clear() {
        size = 0; s[0] = 0;
    }
    inline void reverse() {
        for( int i = 0 ; i < size >> 1 ; ++ i ) swap( s[i], s[size - i - 1] );
    }
    BigInteger( long long num = 0 ) { *this = num;     }

    BigInteger operator = ( long long num ) {
        clear();
        do {
            s[size ++] = num % BASE;
            num /= BASE;
        } while( num > 0 );
        return *this;
    }
    BigInteger operator = ( const string &str ) {
        clear();
        int x, len = ( str.length() - 1 ) / WIDTH + 1;
        for( int i = 0 ; i < len ; ++ i ) {
            int end = str.length() - i * WIDTH;
            int start = max( 0, end - WIDTH );
            sscanf( str.substr( start, end - start ).c_str(), "%d", &x );
            s[size ++] = x;
        }
        return *this;
    }
    BigInteger operator + ( const BigInteger &b ) const {
        BigInteger c;
        c.clear();
        for( int i = 0, g = 0 ; ; ++ i ) {
            if( g == 0 && i >= size && i >= b.size ) break;
            int x = g;
            if( i < size ) x += s[i];
            if( i < b.size ) x += b.s[i];
            c.s[c.size ++] = x % BASE;
            g = x / BASE;
        }
        c.killzero();
        return c;
    }
    BigInteger operator - ( const BigInteger &b ) const {
        BigInteger c;
        c.clear();
        for( int i = 0, g = 0 ; ; ++ i ) {
            if( g == 0 && i >= size && i >= b.size ) break;
            int x = -g;
            if( i < size ) x += s[i];
            if( i < b.size ) x -= b.s[i];
            g = 0;
            while( x < 0 ) { x += BASE ; ++ g;    }
            c.s[c.size ++] = x;
        }
        while( c.s[c.size - 1] == 0 && c.size > 1 ) c.size --;
        return c;
    }
    BigInteger operator * ( const BigInteger &b ) const {
        BigInteger c;
        c.clear();
        for( int i = 0 ; i < b.size ; ++ i ) {
            BigInteger t;
            t.clear();
            if( b.s[i] == 0 ) continue;
            for( int j = 0, g = 0 ; j < size || g != 0; ++ j ) {
                int x = g;
                if( j < size ) x += b.s[i] * s[j];
                t.s[t.size ++] = x % BASE;
                g = x / BASE;
            }
            BigInteger tmp;
            tmp.clear();
            for( int j = 0, g = 0 ; ; ++ j ) {
                if( g == 0 && j >= t.size + i && j >= c.size ) break;
                int x = g;
                if( j < c.size ) x += c.s[j];
                if( j >= i && j < t.size + i ) x += t.s[j - i];
                tmp.s[tmp.size ++] = x % BASE;
                g = x / BASE;
            }
            c = tmp;
        }
        c.killzero();
        return c;
    }
    BigInteger operator / ( const BigInteger &b ) const {
        BigInteger c, t;
        c.clear(); t.clear();
        for( int i = size - 1 ; i >= 0 ; -- i ) {
            t = t * BASE + s[i];
            int x = 0;
            while( b <= t ) { t -= b; x ++ ; }
            c.s[c.size ++] = x;
        }
        c.reverse();
        c.killzero();
        return c;
    }
    BigInteger operator % ( const BigInteger &b ) const {
        BigInteger c;
        c.clear();
        for( int i = size - 1 ; i >= 0 ; -- i ) {
            c = c * BASE + s[i];
            while( b <= c ) c -= b;
        }
        c.killzero();
        return c;
    }

    BigInteger operator += ( const BigInteger &b ) {
        *this = *this + b;
        return *this;
    }
    BigInteger operator -= ( const BigInteger &b ) {
        *this = *this - b;
        return *this;
    }
    BigInteger operator *= ( const BigInteger &b ) {
        *this = *this * b;
        return *this;
    }
    BigInteger operator /= ( const BigInteger &b ) {
        *this = *this / b;
        return *this;
    }
    BigInteger operator %= ( const BigInteger &b ) {
        *this = *this % b;
        return *this;
    }

    BigInteger operator ++ () {
        *this = *this + 1;
        return *this;
    }
    BigInteger operator ++ (int) {
        BigInteger old = *this;
        ++ (*this);
        return old;
    }
    BigInteger operator -- () {
        *this = *this - 1;
        return *this;
    }
    BigInteger operator -- (int) {
        BigInteger old = *this;
        ++ (*this);
        return old;
    } 

    bool operator < ( const BigInteger &b ) const {
        if( size != b.size ) return size < b.size;
        for( int i = size - 1 ; i >= 0 ; -- i )
            if( s[i] != b.s[i] ) return s[i] < b.s[i];
        return false;
    }
    bool operator > ( const BigInteger &b ) const { return b < *this; }
    bool operator <= ( const BigInteger &b ) const { return !( b < *this ); }
    bool operator >= ( const BigInteger &b ) const { return !( *this < b ); }
    bool operator != ( const BigInteger &b ) const { return b < *this || *this < b ;}
    bool operator == ( const BigInteger &b ) const { return !( b < *this ) && !( *this < b ); }
    void read() {
        clear();
        char s1[11000];
        scanf( "%s", s1 );
        int x, len_of_s1 = strlen( s1 ), rem = len_of_s1 % WIDTH , point = 1 - WIDTH;
        for( int i = len_of_s1 - 1 ; i >= rem  ; -- i ) {
            x = x * 10 + s1[i + point] - '0'; point += 2;
            if( ( len_of_s1 - i ) % WIDTH == 0 ) { s[size ++] = x; x = 0; point = 1 - WIDTH; }
        }
        if( rem != 0 ){
            int x = 0;
            for( int i = 0 ; i < rem ; ++ i ) x = x * 10 + s1[i] -'0';
            s[size ++] = x;
        }
    }
    void writeln() {
        printf( "%d", s[size - 1] );
        for(int i = size - 2 ; i >= 0 ; -- i ) printf( "%04d", s[i] );
        putchar( '\n' );
    }
};
BigInteger A, B;
int main(){
    A.read(); B.read();
    BigInteger add = A + B, cut = max( A, B ) - min( A, B ), mul = A * B, div = A / B, rem = A % B;
    add.writeln(); cut.writeln(); mul.writeln(); div.writeln(); rem.writeln();
    return 0;
}

为什么只有七十分？思考之后才发现，其实在做高精度除法的时候，就可以顺带着输出余数，这样才能真好踩着时间点过。

运行截图：

代码：（主程序）

BigInteger A, B;
int main(){
    A.read(); B.read();
    BigInteger add = A + B, cut = max( A, B ) - min( A, B ), mul = A * B;
    add.writeln(); cut.writeln(); mul.writeln();
        BigInteger c, t;
        c.clear(); t.clear();
        for( int i = A.size - 1 ; i >= 0 ; -- i ) {
            t = t * BigInteger::BASE + A.s[i];
            int x = 0;
            while( B <= t ) { t = t - B; x ++ ; }
            c.s[c.size ++] = x;
        }
        for( int i = 0 ; i <= ( c.size - 1 ) >> 1 ; ++ i ) swap( c.s[i], c.s[c.size - i - 1] );
        while( c.s[c.size - 1] == 0 && c.size > 1 ) c.size --;
    c.writeln(); t.writeln();
    return 0;
}