Codeforces 173E Camping Groups hash

Camping Groups

通过分析我们能发现， 如果 i 以前放完了， i 以后随便放的话必定有解。

所以我们只要找出第一个比原串大的地方就好啦， 用hash来判断是不是回文。

#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 5e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;}


int d, n;
char s[N];

LL hs1[N], hs2[N], Pow[N], fPow[N];

LL power(LL a, LL b) {
    LL ans = 1;
    while(b) {
        if(b & 1) ans = ans * a % mod;
        a = a * a % mod; b >>= 1;
    }
    return ans;
}

const int B = 23333;
const int fB = power(B, mod - 2);

LL getHash1(int L, int R) {
    return (hs1[R] - hs1[L - 1] * Pow[R - L + 1] % mod + mod) % mod;
}

LL getHash2(int L, int R) {
    return (hs2[R] - hs2[L - 1] + mod) % mod * fPow[L] % mod;
}

bool check(int L1, int R1, int L2, int R2) {
    if(L1 <= 0) return true;
    return getHash1(L1, R1) != getHash2(L2, R2);
}

int main() {
    for(int i = Pow[0] = 1; i < N; i++) Pow[i] = Pow[i - 1] * B % mod;
    for(int i = fPow[0] = 1; i < N; i++) fPow[i] = fPow[i - 1] * fB % mod;
    scanf("%d%s", &d, s + 1);
    n = strlen(s + 1);
    if(d == 1) return puts("Impossible"), 0;
    int leno, lene;
    if(d & 1) {
        lene = d / 2 + 1;
        leno = d / 2 + 1;
    } else {
        lene = d / 2;
        leno = d / 2 + 1;
    }
    int where = 0;
    for(int i = 1; i <= n; i++) {
        bool flag = false;
        for(int j = s[i] + 1; j <= 'z'; j++) {
            hs1[i] = (hs1[i - 1] * B % mod + j) % mod;
            hs2[i] = (hs2[i - 1] + Pow[i] * j % mod) % mod;
            if(check(i - 2 * lene + 1, i - lene, i - lene + 1, i) && check(i - 2 * leno + 2, i - leno + 1, i - leno + 1, i)) {
                flag = true;
                break;
            }
        }
        if(flag) where = i;
        hs1[i] = (hs1[i - 1] * B % mod + s[i]) % mod;
        hs2[i] = (hs2[i - 1] + Pow[i] * s[i] % mod) % mod;
        if(!check(i - 2 * lene + 1, i - lene, i - lene + 1, i) || !check(i - 2 * leno + 2, i - leno + 1, i - leno + 1, i)) {
            break;
        }
    }
    if(!where) puts("Impossible");
    else {
        for(int i = 1; i <= n; i++) {
            if(i >= where) {
                int down = i == where ? s[i] + 1 : 'a';
                for(int j = down; j <= 'z'; j++) {
                    hs1[i] = (hs1[i - 1] * B % mod + j) % mod;
                    hs2[i] = (hs2[i - 1] + Pow[i] * j % mod) % mod;
                    if(check(i - 2 * lene + 1, i - lene, i - lene + 1, i) && check(i - 2 * leno + 2, i - leno + 1, i - leno + 1, i)) {
                        s[i] = j;
                        break;
                    }
                }
            }
            hs1[i] = (hs1[i - 1] * B % mod + s[i]) % mod;
            hs2[i] = (hs2[i - 1] + Pow[i] * s[i] % mod) % mod;
        }
        puts(s + 1);
    }
    return 0;
}

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