/* Problem Link : http://www.lightoj.com/volume_showproblem.php?problem=1258 */ #include <bits/stdc++.h> #define pii pair <int,int> #define sc scanf #define pf printf #define Pi 2*acos(0.0) #define ms(a,b) memset(a, b, sizeof(a)) #define pb(a) push_back(a) #define MP make_pair #define oo 1<<29 #define dd double #define ll long long #define EPS 10E-10 #define ff first #define ss second #define MAX 1000007 #define CIN ios_base::sync_with_stdio(0) #define SZ(a) (int)a.size() #define getint(a) scanf("%d",&a) #define getint2(a,b) scanf("%d%d",&a,&b) #define getint3(a,b,c) scanf("%d%d%d",&a,&b,&c) #define loop(i,n) for(int i=0;i<n;i++) #define TEST_CASE(t) for(int z=1;z<=t;z++) #define PRINT_CASE printf("Case %d: ",z) #define all(a) a.begin(),a.end() #define intlim 2147483648 #define inf 1000000 #define rtintlim 46340 #define llim 9223372036854775808 #define rtllim 3037000499 #define ull unsigned long long #define I int using namespace std; /* Bits operation */ int Set(int n,int pos) { return n = n | 1<<pos;} bool check(int n,int pos) { return n & 1<<pos;} int Reset(int n, int pos) { return n=n & ~(1<<pos);} /*----------------*/ int lps[MAX]; void computePrefixFunction(string& P) { lps[0]=-1; int k=-1; int n=SZ(P); for(int i=1;i<n;i++) { while(k>-1 && P[i]!=P[k+1]) k=lps[k]; if(P[i]==P[k+1]) k++; lps[i]=k; } } int KMP(string& T, string& P) { int n=SZ(T); int m=SZ(P); int k=-1; computePrefixFunction(P); int cnt=0; for(int i=0;i<n;i++) { while(k>-1 && T[i]!=P[k+1]) k=lps[k]; if(T[i]==P[k+1]) k++; if(i==n-1) cnt=k; } return cnt+1; } int main() { ///freopen("in.txt","r",stdin); ///freopen("out.txt","w",stdout); int t; getint(t); TEST_CASE(t) { string txt,pattarn; cin>>txt; pattarn=txt; reverse(all(pattarn)); int n=KMP(txt,pattarn); int m=SZ(txt); PRINT_CASE; cout<<m+m-n<<endl; } return 0; }
what does kmp do in this problem and why you make your ans by doing (m+m-n)?
Here KMP is used to determine longest common pattern in the given string and it’s reverse string. then if we add the reverse of a string in the end of this string then it will be a palindrome. then the common part is subtracted from the length.