PAT(Advanced) 1132 Cut Integer C++实现-优快云博客

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PAT(Advanced) 1132 Cut Integer C++实现

题目链接

[1132 Cut Integer[(https://pintia.cn/problem-sets/994805342720868352/problems/994805347145859072)

题目大意

给定K位的十进制数值Z，根据其前后两部分分为K/2的两个数值A和B，判断Z能否被A和B的乘积整除，若能则输出Yes，否则输出No

算法思路

对Z除基取余法取后K/2位得到B，操作后的Z即为A，进行判断即可，注意B为0的可能会导致取余0浮点错误，需特殊处理（B为0时Z必然不能0被整除，返回false即可）。

AC代码

/*
author : eclipse
email  : eclipsecs@qq.com
time   : Tue Feb 02 10:54:52 2021
*/
#include<bits/stdc++.h>
using namespace std;

int calculateLength(int number) {
    int length = 0;
    while (number) {
        length++;
        number /= 10;
    }

    return length;
}

bool divided(int Z) {
    int A = Z;
    int length = calculateLength(A);
    int B = 0;
    int base = 1;
    for (int i = 0; i < length / 2; i++) {
        B += (A % 10) * base;
        base *= 10;
        A /= 10;
    }
    return B == 0 ? false: Z % (A * B) == 0;
}

int main(int argc, char const *argv[]) {
    int N;
    scanf("%d", &N);
    while (N--) {
        int Z;
        scanf("%d", &Z);
        printf("%s\n", divided(Z) ? "Yes" : "No");
    }
    return 0;
}