【PAT甲级】1132 Cut Integer（20 分）_1132 cut integer (20分) 柳婼-优快云博客

本文介绍了一种算法，用于判断一个整数是否能通过切割成两个子整数，并使得原整数能被这两个子整数的乘积整除。通过输入一系列整数，程序能够输出每个整数是否符合这一有趣特性。

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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 <231). 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:

Sample Output:

Yes
No
No

思路：利用sscanf(str,"%d",&x)来切分数字模拟

代码：

#include <iostream>
#include <sstream>
#include <string>
#include <cstdio>
#include <queue>
#include <cstring>
#include <map>
#include <cstdlib>
#include <stack>
#include <cmath>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define rep(i,a,n) for(int i=a;i<n;i++)
#define mem(a,n) memset(a,n,sizeof(a))
#define DBGS() cout<<"START\n"
#define DBGE() cout<<"END\n"
const int N = 256+5;
const ll INF = 0x3f3f3f3f;
const double eps=1e-4;

int main() {
    int n;
    scanf("%d",&n);
    char str[20];
    int all,l,r;
    for(int i=0; i<n; i++) {
        scanf("%s",str);
        sscanf(str,"%d",&all);
        int len=strlen(str)/2;
        sscanf(str+len,"%d",&r);
        str[len]='\0';
        sscanf(str,"%d",&l);
        if(!l||!r) {
            puts("No");
        } else {
            if(all%(l*r)==0)
                puts("Yes");
            else
                puts("No");
        }
    }
    return 0;
}