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);
    }
}
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