洛谷P1601 A+B problem 洛谷P1303 A*B problem

寫在前面：

這兩道題我採用了C++和java兩種不同的方法去寫，個人覺得還是Java寫高精度模擬又快又好，當然對於一個打算搞acm的蒟蒻，兩種方法還是要學會，才能幹！有時候出題人，故意爲難，用java會卡時，這篇只是記錄，大家要是想知道具體的方法可以參照這篇博客：

大整數的操作—階乘，加法，乘法-----比對C++與java

大精度和與積

題目鏈接：
洛谷P1601 A+B problem
洛谷P1303 A*B problem

首先先寫A+B

java：

import java.util.*;
import java.math.*;

public class Main {
    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in);
        BigInteger a, b;
        a = cin.nextBigInteger();
        b = cin.nextBigInteger();
        System.out.println(a.add(b));
        cin.close();
    }
}

C++模擬豎式運算：

#include <bits/stdc++.h>
using namespace std;
const int maximum = 10e3;
int sum[maximum];
char a[maximum], b[maximum];
int main()
{
    memset(sum, 0, sizeof(sum));
    memset(a, '0', sizeof(a));
    memset(b, '0', sizeof(b));
    scanf("%s%s", a, b);
    char transposedA[maximum];
    char transposedB[maximum];
    memset(transposedA, '0', sizeof(transposedA));
    memset(transposedB, '0', sizeof(transposedB));
    for (int i = strlen(a) - 1, j = 0; i >= 0; --i, ++j)
        transposedA[j] = a[i];    //  將a數組轉置過來
    for (int i = strlen(b) - 1, j = 0; i >= 0; --i, ++j)
        transposedB[j] = b[i];    //  將b數組轉置過來
    int len = max(strlen(a), strlen(b));
    for (int i = 0; i < len; i++)
        sum[i] = transposedA[i] - '0' + transposedB[i] - '0';    //  模擬豎式加法
    int up = 0;
    int k = 0;
    while (k < len)    //  模擬進位
    {
        sum[k] += up;
        up = sum[k] / 10;
        sum[k] %= 10;
        k++;
    }
    while (up)
    {
        sum[k++] = up % 10;
        up /= 10;
    }
    for (int i = k - 1; i >= 0; --i)
        printf("%d", sum[i]);
    cout << endl;
    return 0;
}

A*B problem

java

import java.util.*;
import java.math.*;

public class BigInter {
	public static void main(String[] args) {
		Scanner cin = new Scanner(System.in);
		BigInteger a, b;
		a = cin.nextBigInteger();
		b = cin.nextBigInteger();
		System.out.println(a.multiply(b));
		cin.close();
	}
}

C++模擬豎式乘法：

#include <bits/stdc++.h>
using namespace std;
const int maximum = 10e4;
int sum[maximum];
char a[maximum], b[maximum];
int main()
{
    memset(sum, 0, sizeof(sum));
    memset(a, '\0', sizeof(a));
    memset(b, '\0', sizeof(b));
    scanf("%s%s", a, b);
    char transposedA[maximum];
    char transposedB[maximum];
    memset(transposedA, '\0', sizeof(transposedA));
    memset(transposedB, '\0', sizeof(transposedB));
    for (int i = strlen(a) - 1, j = 0; i >= 0; --i, ++j)
        transposedA[j] = a[i];     //  將a數組轉置過來
    for (int i = strlen(b) - 1, j = 0; i >= 0; --i, ++j)
        transposedB[j] = b[i];     //  將b數組轉置過來
    for (int i = 0; i < strlen(transposedA); ++i)
    {
        for (int j = 0; j < strlen(transposedB); ++j)
            sum[i + j] += (transposedA[i] - '0') * (transposedB[j] - '0');    //  模擬豎式乘法
    }
    int len = strlen(transposedA) + strlen(transposedB) - 1;
    int up = 0;
    int k = 0;
    while (k < len)    //  模擬進位
    {
        sum[k] += up;
        up = sum[k] / 10;
        sum[k] %= 10;
        k++;
    }
    while (up)
    {
        sum[k++] = up % 10;
        up /= 10;
    }
    if (a[0]=='0'||b[0]=='0')  //  判斷有無爲0的乘數
    {
        cout << "0" << endl;
    }
    else
    {
        for (int i = k - 1; i >= 0; --i)
            printf("%d", sum[i]);
        cout << endl;
    }
    return 0;
}