PAT 甲级 1132 Cut Integer
https://pintia.cn/problem-sets/994805342720868352/problems/994805347145859072
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 <). 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
代码:
#include <bits/stdc++.h> using namespace std; const int maxn = 1e5 + 10; char s[maxn]; void itoa(int x) { if(x == 0) { s[0] = '0'; s[1] = 0; return ; } stack<int> st; while(x) { st.push(x % 10); x = x / 10; } int sz = 0; while(!st.empty()) { s[sz++] = (char)(st.top() + '0'); s[sz] = 0; st.pop(); } } int main() { int T; scanf("%d", &T); while(T --) { int n; int num1 = 0, num2 = 0; scanf("%d", &n); itoa(n); //printf("%s\n", s); int len = strlen(s); for(int i = 0; i < len / 2; i ++) num1 = (s[i] - '0') + num1 * 10; for(int i = len / 2; i < len; i ++) num2 = (s[i] - '0') + num2 * 10; if(num1 * num2 == 0) printf("No\n"); else { if(n % (num1 * num2) == 0) printf("Yes\n"); else printf("No\n"); } //printf("%d %d", num1, num2); } return 0; }