C. Distinct Substrings
大意: 给定串$s$, 字符集$m$, 对于每个字符$c$, 求$s$末尾添加字符$c$后本质不同子串增加多少.
exkmp求出每个前缀与后缀匹配的最大长度, 统计一下贡献即可
#include <iostream> #include <sstream> #include <algorithm> #include <cstdio> #include <cmath> #include <set> #include <map> #include <queue> #include <string> #include <cstring> #include <bitset> #include <functional> #include <random> #define REP(i,a,n) for(int i=a;i<=n;++i) #define PER(i,a,n) for(int i=n;i>=a;--i) #define hr putchar(10) #define pb push_back #define lc (o<<1) #define rc (lc|1) #define mid ((l+r)>>1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl '\n' #define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;}) using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 1e9+7, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;} //head const int N = 1e6+50; int n,m,a[N],z[N],po[N],h[N],b[N]; void init(int *s, int *z, int n) { int mx=0,l=0; REP(i,1,n-1) { z[i] = i<mx?min(mx-i,z[i-l]):0; while (i+z[i]<n&&s[z[i]]==s[i+z[i]]) ++z[i]; if (i+z[i]>mx) mx=i+z[i],l=i; } } int main() { po[0] = 1; REP(i,1,N-1) po[i] = po[i-1]*3ll%P; while (~scanf("%d%d", &n, &m)) { REP(i,1,m) h[i] = -1; REP(i,1,n) { scanf("%d",a+i); h[a[i]] = 0; } REP(i,1,n) b[i]=a[n-i+1]; init(b+1,z+1,n); REP(i,1,n-1) h[a[i+1]] = max(h[a[i+1]],z[n-i+1]); ll ans = 0; REP(i,1,m) ans ^= (ll)(n-h[i])*po[i]%P; printf("%lld\n", ans); } }