标签：while specific cpp not code ons ++ http contain
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.
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.
For each case, print a single line Yes if it is such a number, or No if not.
3
167334
2333
12345678
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;
}
标签：while specific cpp not code ons ++ http contain
原文地址：https://www.cnblogs.com/zlrrrr/p/9420583.html
