A1132. Cut Integer

Cutting an integer means to cut a K digits lone integer Z into two integers of (K/2) digits long integers A and B. For example, after cutting Z = 167334, we have A = 167 and B = 334. It is interesting to see that Z can be devided by the product of A and B, as 167334 / (167 × 334) = 3. Given an integer Z, you are supposed to test if it is such an integer.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤ 20). Then N lines follow, each gives an integer Z (10 ≤ Z <2^31). It is guaranteed that the number of digits of Z is an even number.

Output Specification:

For each case, print a single line Yes if it is such a number, or No if not.

Sample Input:

3
167334
2333
12345678

Sample Output:

Yes
No
No

#include<iostream>
#include<cstdio>
using namespace std;
long long Z;
int num2Cut(long long Z){
    long long temp[], bit, Z2 = Z;
    int pt = ;
    do{
        bit = Z % ;
        Z = Z / ;
        temp[pt++] = bit;
    }while(Z != );
    long long a = , b = ;
    for(int i = pt - ; i >= pt / ; i--){
        a = a *  + temp[i];
    }
    for(int i = pt /  - ; i >= ; i--){
        b = b *  + temp[i];
    }
    if(a*b == )
        return ;
    if(Z2 % (a*b) == )
        return ;
    else return ;
}
int main(){
    int N;
    scanf("%d", &N);
    for(int i = ; i < N; i++){
        scanf("%lld", &Z);
        int tag = num2Cut(Z);
        if(tag == )
            printf("No\n");
        else printf("Yes\n");
    }
    cin >> N;
    return ;
}

总结：
1、题意：将一个位数为偶数的数字分成两半，看看这两半的乘积能不能被原数整除。
2、注意依次取一个数的各个位，得到的数组中是倒着的，由高位到低位。
2、在验证能否整除时，要注意被除数为0的情况。如1000被分成10和00.