Idea: If you want to truncate an even integer into two new integers, you need to find the total length of this number first, assuming the length is 2n. Then the first n is the first integer, which can be obtained by dividing the original integer by the n power of 10; The last n bit is the second integer, which can be obtained by dividing the original integer by the n power of 10.

Reference code:

# include & ltstring.h & gt

# include & ltstdio.h & gt

# include & ltmath.h & gt

int fun(int a){

int n = 0；

while(a){

n++；

a/= 10；

}

Returns n/2;

}

int main()

{

int a，n，t；

scanf("%d "，& ampa)；

n = fun(a)；

t=(int)pow( 10，n)；

printf("%d %d "，a/t，a % t)；

Returns 0;

}

/*

Output:

12345678

1234 5678

*/