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.
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; }