A. Mike and Fax
Time Limit: 20 Sec Memory Limit: 256 MB
题目连接
http://codeforces.com/contest/548/problem/A
Description
While Mike was walking in the subway, all the stuff in his back-bag dropped on the ground. There were several fax messages among them. He concatenated these strings in some order and now he has string s.
![](http://codeforces.com/predownloaded/45/8b/458b55d513eb7e8e9e408f9edda4618def368721.png)
He is not sure if this is his own back-bag or someone else's. He remembered that there were exactly k messages in his own bag, each was a palindrome string and all those strings had the same length.
He asked you to help him and tell him if he has worn his own back-bag. Check if the given string s is a concatenation of k palindromes of the same length.
Input
The first line of input contains string s containing lowercase English letters (1 ≤ |s| ≤ 1000).
The second line contains integer k (1 ≤ k ≤ 1000).
Output
Print "YES"(without quotes) if he has worn his own back-bag or "NO"(without quotes) otherwise.
Sample Input
saba
2
Sample Output
NO
HINT
题意
给你个字符串,告诉你这个字符串是由k个相同长度的回文串构成的,判断是否正确
题解:
数据范围很小,暴力判断就好了
代码:
//qscqesze #include <cstdio> #include <cmath> #include <cstring> #include <ctime> #include <iostream> #include <algorithm> #include <set> #include <vector> #include <sstream> #include <queue> #include <typeinfo> #include <fstream> #include <map> #include <stack> typedef long long ll; using namespace std; //freopen("D.in","r",stdin); //freopen("D.out","w",stdout); #define sspeed ios_base::sync_with_stdio(0);cin.tie(0) #define test freopen("test.txt","r",stdin) #define maxn 200001 #define mod 1000000007 #define eps 1e-9 int Num; char CH[20]; //const int inf=0x7fffffff; //нчоч╢С const int inf=0x3f3f3f3f; inline ll read() { ll x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } inline void P(int x) { Num=0;if(!x){putchar('0');puts("");return;} while(x>0)CH[++Num]=x%10,x/=10; while(Num)putchar(CH[Num--]+48); puts(""); } //************************************************************************************** string s; int k; int main() { //test; cin>>s>>k; if(s.size()%k!=0) { cout<<"NO"<<endl; return 0; } int len=s.size()/k; for(int i=0;i<k;i++) { for(int j=0;j<len;j++) { if(s[i*len+j]!=s[i*len+len-j-1]) { cout<<"NO"<<endl; return 0; } } } cout<<"YES"<<endl; }