题目

Description

求两个不超过50位的非负数的积。

Input

有两行，每行是一个不超过50位的非负整数，没有多余的前导0

Output

一行，即相乘后的结果。结果里不能有多余的前导0，即如果结果是0342，则输出342.

Sample Input

12345678900
98765432100

Sample Output

1219326311126352690000
题解

>高精度算法<
代码

/*
By:OhYee
Github:OhYee
Blog:http://www.oyohyee.com/
Email:oyohyee@oyohyee.com

かしこいかわいい？
エリーチカ！
要写出来Хорошо的代码哦~
*/
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <functional>
using namespace std;

#define REP(n) for(int o=0;o<n;o++)
#define Max(a,b) (a>b?a:b)
#define Min(a,b) (a<b?a:b)
const int maxn = 10000;

class bigNumber {
private:
    int num[maxn + 1];
public:
    bigNumber() {
        init();
    }
    void init() {
        memset(num,0,sizeof(num));
    }
    bigNumber operator = (const char *str) {
        init();
        num[0] = strlen(str);
        REP(num[0])num[num[0] - o] = str[o] - '0';
        return *this;
    }
    bigNumber operator = (const bigNumber& rhs) {
        init();
        REP(rhs.num[0] + 1)num[o] = rhs.num[o];
        return *this;
    }
    bigNumber operator = (long long rhs) {
        init();
        int i = 1;
        while(rhs) {
            num[i] = rhs % 10;
            rhs /= 10;
            i++;
        }
        num[0] = i - 1;
        return *this;
    }
    bool operator < (const bigNumber rhs)const {
        if(num[0] != rhs.num[0])return (num[0] < rhs.num[0]);
        REP(num[0]) {
            int temp = num[0] - o;
            if(num[temp] != rhs.num[temp])return (num[temp] < rhs.num[temp]);
        }
        return 0;
    }
    bool operator > (const bigNumber rhs)const {
        if(num[0] != rhs.num[0])return (num[0] > rhs.num[0]);
        REP(num[0]) {
            int temp = num[0] - o;
            if(num[temp] != rhs.num[temp])return (num[temp] > rhs.num[temp]);
        }
        return 0;
    }
    bool operator == (const bigNumber rhs)const {
        return !(*this > rhs || *this < rhs);
    }
    bool operator <= (const bigNumber rhs)const {
        return *this < rhs || *this == rhs;
    }
    bigNumber operator + (const bigNumber rhs)const {
        bigNumber temp;
        int len;
        len = num[0] > rhs.num[0] ? num[0] : rhs.num[0];
        len++;
        REP(len) {
            temp.num[o + 1] += num[o + 1] + rhs.num[o + 1];
            temp.num[o + 2] += temp.num[o + 1] / 10;
            temp.num[o + 1] %= 10;
        }
        REP(len) {
            if(temp.num[len - o] != 0) {
                temp.num[0] = len - o;
                break;
            }
        }
        return temp;
    }
    bigNumber operator + (const long long rhs)const {
        bigNumber temp1,temp2;
        temp1 = *this;
        temp2 = rhs;
        return temp1 + temp2;
    }
    bigNumber operator * (const bigNumber rhs)const {
        bigNumber temp;
        int len;
        len = num[0] + rhs.num[0];
        //len++;
        for(int i = 1;i <= num[0];i++) {
            for(int j = 1;j <= rhs.num[0];j++) {
                temp.num[i + j - 1] += num[i] * rhs.num[j];
                temp.num[i + j] += temp.num[i + j - 1] / 10;
                temp.num[i + j - 1] %= 10;
            }
        }
        REP(len) {
            if(temp.num[len - o] != 0) {
                temp.num[0] = len - o;
                break;
            }
        }
        return temp;
    }
    bigNumber operator * (const long long rhs)const {
        bigNumber temp1,temp2;
        temp1 = *this;
        temp2 = rhs;
        return temp1 * temp2;
    }
    bigNumber operator - (const bigNumber rhs)const {
        bigNumber temp,a,b;
        temp = Max(*this,rhs);
        b = Min(*this,rhs);
        a = temp;
        temp.init();
        int len = a.num[0];
        REP(len) {
            temp.num[o + 1] += 10 + a.num[o + 1] - b.num[o + 1];
            temp.num[o + 2]--;
            temp.num[o + 2] += temp.num[o + 1] / 10;
            temp.num[o + 1] %= 10;
        }
        REP(len) {
            if(temp.num[len - o] != 0) {
                temp.num[0] = len - o;
                break;
            }
        }
        return temp;
    }
    bigNumber operator - (const long long rhs)const {
        bigNumber temp1,temp2;
        temp1 = *this;
        temp2 = rhs;
        return temp1 - temp2;
    }
    bigNumber operator / (const bigNumber rhs)const {
        bigNumber a;
        int it = num[0];

        bigNumber d;
        bigNumber c;

        while(it > 0) {
            a = (d * 10) + num[it];
            c = c * 10;

            int t;
            REP(9) {
                if(a < rhs * (o + 1)) {
                    t = o;
                    break;
                }
                t = 9;
            }
            c = c + t;
            d = a - rhs*t;
            it--;
        }
        return c;
    }
    bigNumber operator / (const long long rhs)const {
        bigNumber temp1,temp2;
        temp1 = *this;
        temp2 = rhs;
        return temp1 / temp2;
    }
    bigNumber operator % (const bigNumber rhs)const {
        bigNumber a;
        int it = num[0];

        bigNumber d;
        bigNumber c;



        while(it > 0) {

            a = (d * 10) + num[it];
            c = c * 10;

            int t;
            REP(9) {
                if(a < rhs * (o + 1)) {
                    t = o;
                    break;
                }
                t = 9;
            }
            c = c + t;
            d = a - rhs*t;
            it--;
        }
        return d;
    }
    bigNumber operator % (const long long rhs)const {
        bigNumber temp1,temp2;
        temp1 = *this;
        temp2 = rhs;
        return temp1 % temp2;
    }
    void p() {
        if(num[0] == 0)
            printf("0");
        REP(num[0]) {
            printf("%d",num[num[0] - o]);
        }
    }
};

bigNumber a,b,c;
char aa[maxn],bb[maxn];
bool Do() {
    
    if(scanf("\n%s\n%s",aa,bb)==EOF)
        return false;

    a = aa;
    b = bb;
    c = a * b;
    c.p();

    putchar('\n');
    return true;
}

int main() {
    while(Do());
    return 0;
}
AOJ 770.大整数乘法
