A. Cut Integer (20)

時(shí)間限制 400 ms
內(nèi)存限制 65536 kB
代碼長(zhǎng)度限制 8000 B
判題程序 Standard 作者 CHEN, Yue

Cutting an integer means to cut a K digits long 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 x 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: 3 167334 2333 12345678 Sample Output: Yes No No思路+題意：求給定的一個(gè)偶數(shù)為的整數(shù)，將這個(gè)數(shù)分成兩個(gè)長(zhǎng)度相同的數(shù)相乘被原來(lái)的數(shù)取余是否為0；直接求解，將數(shù)分兩半，進(jìn)行相乘求解。注意這里有可能是123000，其結(jié)果就是123*000=0； 這里會(huì)出現(xiàn)浮點(diǎn)錯(cuò)誤，答案就為No。 代碼： #include <iostream> #include <cmath> using namespace std; int main() {int n, num;scanf("%d", &n);for (int i = 0; i < n; i++) {scanf("%d", &num);int len = 0, tempnum = num;while (tempnum != 0) {tempnum = tempnum / 10;len++;}int d = pow(10, len / 2);int a = num % d, b = num / d;if (a * b != 0 && num % (a * b) == 0)printf("Yes\n");elseprintf("No\n");}return 0; }

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