Question

我正在尝试将两个64位整数相乘并得到错误 - 当我尝试将产品存储在uint512_t数据类型中时，uint512_t未在此范围内声明。是否有可用于存储此类巨大值的替代数据类型？我的数组包含我想要乘以的数字的数字。

#include <cstdint>
#include <iostream>
#include <stdint.h>
using namespace std;

int multiply(int x, int y, int carry)
{
    int product;
    product = x * y + carry;
    return product;
}

int add(int multiplier, int product_current, int product_new)
{
    product_current = product_current + multiplier * product_new;
    return product_current;
}

int main()
{
    int a[64] = { 3, 1, 4, 1, 5, 9, 2, 6, 5, 3, 5, 8, 9, 7, 9, 3, 2, 3, 8, 4, 6, 2, 6, 4, 3, 3, 8, 3, 2, 7, 9, 5, 0, 2, 8, 8, 4, 1, 9, 7, 1, 6, 9, 3, 9, 9, 3, 7, 5, 1, 0, 5, 8, 2, 0, 9, 7, 4, 9, 4, 4, 5, 9, 2 };
    int b[64] = { 2, 7, 1, 8, 2, 8, 1, 8, 2, 8, 4, 5, 9, 0, 4, 5, 2, 3, 5, 3, 6, 0, 2, 8, 7, 4, 7, 1, 3, 5, 2, 6, 6, 2, 4, 9, 7, 7, 5, 7, 2, 4, 7, 0, 9, 3, 6, 9, 9, 9, 5, 9, 5, 7, 4, 9, 6, 6, 9, 6, 7, 6, 2, 7 };
    int carryin = 0;
    uint512_t temp_result = 0;
    uint512_t temp_product = 0;
    int temp_carry = 0;
    uint512_t product_acch = 0;
    uint512_t product_accr = 0;
    uint512_t mul = 1;
    uint512_t mul2 = 1;
    for (int i = 3; i >= 0; i--) {
        carryin = 0;
        product_acch = 0;
        mul = 1;
        for (int j = 3; j >= 0; j--) {
            temp_result = multiply(a[j], b[i], carryin);
            temp_product = temp_result % 10;
            temp_carry = temp_result / 10;
            product_acch = add(mul, product_acch, temp_product);
            mul = mul * 10;
            carryin = temp_carry;
            if (carryin != 0 && j == 0) {
                product_acch = product_acch + mul * carryin;
            }
        }
        product_accr = add(mul2, product_accr, product_acch);

        cout << product_accr << endl;
        mul2 = mul2 * 10;
    }

    cout << product_accr;
    return 0;
}

Answer 1

使用Boost：

#include <boost/multiprecision/cpp_int.hpp>

using namespace boost::multiprecision;

int512_t x;
uint512_t y;

如何在C ++中定义512位整数？

1 个答案: