C++ 将两个数字合二为一。示例：123 和 456 变成 123456
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Combine two numbers into one. Example: 123 and 456 become 123456
提问by jiake
In C++, how do I combine (note: not add) two integers into one big integer?
在 C++ 中，如何将两个整数组合（注意：不是相加）成一个大整数？
For example:
例如：
int1 = 123;
int2 = 456;
Is there a function to take the two numbers and turn intCombined into 123456?
有没有函数可以将两个数字转换为 intCombined 为 123456？
EDIT:
编辑：
My bad for not explaining clearly. If int2 is 0, then the answer should be 123, not 1230. In actuality though, int1 (the number on the left side) would only have a value if int2 goes over the 32 bit limit. So when int2 is 0, then int1 is 0 (or garbage, i'm not sure).
我不好解释清楚。如果 int2 为 0，那么答案应该是 123，而不是 1230。但实际上，int1（左侧的数字）只有在 int2 超过 32 位限制时才会有值。因此，当 int2 为 0 时，则 int1 为 0（或垃圾，我不确定）。
回答by sth
The power of ten, that you need to multiply the first number with, is the smallest one, that is bigger than the second number:
需要与第一个数字相乘的十的幂是最小的，即大于第二个数字：
int combine(int a, int b) {
int times = 1;
while (times <= b)
times *= 10;
return a*times + b;
}
回答by Jan Johansen
You could convert them into strings, combine them and then convert them back to an int?
您可以将它们转换为字符串，组合它们，然后将它们转换回 int？
回答by R Samuel Klatchko
For each digit in int2, you can multiple int1by 10 and then add int2:
对于 中的每个数字int2，您可以int1乘以 10，然后添加int2：
// merge(123, 0) => 1230
int merge(int int1, int int2)
{
int int2_copy = int2;
do
{
int1 *= 10;
int2_copy /= 10;
} while (int2_copy);
return int1 + int2;
}
You could get rid of the loop using log10and ceil.
您可以使用log10and摆脱循环ceil。
回答by Noel
If the numbers you are trying to combine are positive integers you can use pairing Functions.
如果您尝试组合的数字是正整数，您可以使用配对 Functions。
Pairing function creates a unique number from two. It is also a reversible function.
配对功能从两个创建一个唯一的数字。它也是一个可逆函数。
x,y -> z
z -> x,y
z = (x+y)(x+y+1)/2 + y
Then the reverse is:
那么反过来就是：
w = floor((sqrt(8z+1)-1)/2)
t = (w*w + w)/2
y = z - t
x = w - y
Note. The above is not in any specific language. Just some math...
笔记。以上不是任何特定语言。只是一些数学...
回答by csj
Assuming both ints are non-negative, and int1 goes on the left and int2 goes on the right, you need to figure out how many digits long int2 is, multiply int1 by 10 a bunch of times, and then add them.
假设两个int都是非负的，int1在左边，int2在右边，你需要算出int2的长度是多少，将int1乘以10一堆，然后相加。
unsigned int int1 = blah;
unsigned int int2 = blah;
unsigned int temp = int2;
do
{
temp /= 10;
int1 *= 10;
} while (temp >0)
unsigned int newInt = int1 + int2;
回答by wheaties
You could use stringstream:
您可以使用字符串流：
string Append(int _1, int _2){
stringstream converter;
converter << _1 << _2;
return converter.str();
}
then call atoion the returned string.
然后调用atoi返回的字符串。
回答by Clifford
The following is essentially sth's accepted solution but with the b==0 fix, and the loop replaced with an expression to calculate the scale directly:
以下本质上是 sth 接受的解决方案，但使用 b==0 修复，并且循环替换为表达式以直接计算比例：
#include <math.h>
int combine(int a, int b)
{
int times = 1;
if( b != 0 )
{
times = (int)pow(10.0, (double)((int)log10((double)b)) + 1.0);
}
return a * times + b ;
}
In some circumstances (such as a target with an FPU and a good maths library) the expression might be faster than the loop, but I have not tested that hypothesis.
在某些情况下（例如具有 FPU 和良好数学库的目标），表达式可能比循环更快，但我尚未测试该假设。
回答by Juliano
Another option that works for C too:
另一个也适用于 C 的选项：
#include <stdio.h>
int CombineInt(int int1, int int2)
{
char cResult[32];
sprintf(cResult, "%d%d", int1, int2);
return atoi(cResult);
}
回答by Mehtab Khan
#include <iostream>
using namespace std;
int main()
{
int num1,num2,comb,a,c;
cout << "Enter the 1st numbers" << endl;
cin>>num1;
cout << "Enter the 2st numbers" << endl;
cin>>num2;
a=num2/10;
if(a<=9){
c=num1*100;
comb=c+num2;
cout<<"The combination of the two numbers is "<<comb;
}
else if(a>9&&a<=19){
c=num1*1000;
comb=c+num2;
cout<<"The combination of the two numbers is "<<comb<<endl;
}
else if(a>19&&a<=29){
c=num1*10000;
comb=c+num2;
cout<<"The combination of the two numbers is "<<comb<<endl;
}
return 0;
}
