两个非常大的数的乘积的第一个数字 [英] First digit of the product of two very large numbers

问题描述

有2个大整数，x和y使得z = x * y溢出。
我想计算x的第一个数字。直接这样做是不可能的，因为结果溢出。

我想知道是否有一个特定的技术。

例如：10000000000000000000000000000000000 * 20579725928294522859735727575，这里的第一个数字是2，我可以直接看到，但是，除了这个特殊情况，有没有办法？

  #include< iostream> 
 using namespace std; 
 
 int main（）{
 long long int x，y，z; 
 cin>> x>> y; 
 z = x * y; 
 while（z> 9）
 {
 z / = 10; 
} 
 cout<< z; 
 return 0; 
} 
  

上面的代码无法正常工作。
对于
x = 10000000000000000000000000000000000
y = 20579725928294522859735727575

它将输出5作为z溢出，因此z = 5240626126797720724，这显然是错误的。

您可以用字符串 来执行乘法

，就像我多年前在学校学到的。

 
 123456789987654321 * 998877665544332211 
 
 123456789987654321 
 * 998877665544332211 
 -------------------------------------- 
 123456789987654321 
 123456789987654321 
 246913579975308642 
 246913579975308642 
 370370369962962963 
 370370369962962963 
 493827159950617284 
 493827159950617284 
 617283949938271605 
 617283949938271605 
 740740739925925926 
 740740739925925926 
 864197529913580247 
 864197529913580247 
 987654319901234568 
 987654319901234568 
 1111111109888888889 
 + 1111111109888888889 
 -------------- ------------------------ 
 123318230178465034444583004253633731 
 

There are 2 large integers, x and y such that z=x*y overflows. I would like to compute the first digit of x. Directly doing so isn't possible as the result overflows.

I was wondering if there was a certain technique for this.

Example: 10000000000000000000000000000000000 * 20579725928294522859735727575, here the first digit is 2 as I can see it directly, however, besides this special case, is there a way?

 #include <iostream>
using namespace std;

int main() {
long long int x,y,z;
cin>>x>>y;
z=x*y;
while(z>9)
{
    z/=10;
}
cout<<z;
    return 0;
 }

Above is a code that fails to work. For x=10000000000000000000000000000000000 y=20579725928294522859735727575

It gives output 5 as z overflows, and thus z=5240626126797720724 which is clearly wrong. I am looking to find a way out of this.

解决方案

You can do the multiplication with strings, like I learned in school many years ago

123456789987654321 * 998877665544332211

                    123456789987654321
                  * 998877665544332211
--------------------------------------
                    123456789987654321
                   123456789987654321
                  246913579975308642
                 246913579975308642
                370370369962962963
               370370369962962963
              493827159950617284
             493827159950617284
            617283949938271605
           617283949938271605
          740740739925925926
         740740739925925926
        864197529913580247
       864197529913580247
      987654319901234568
     987654319901234568
   1111111109888888889
+ 1111111109888888889
--------------------------------------
  123318230178465034444583004253633731

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