EOJ 2020.7 月赛 D. 前缀排序 (哈希)

题目：传送门

题意

 思路

官方题解

#include <bits/stdc++.h>
#define LL long long
#define ULL unsigned long long
#define UI unsigned int
#define mem(i, j) memset(i, j, sizeof(i))
#define rep(i, j, k) for(int i = j; i <= k; i++)
#define dep(i, j, k) for(int i = k; i >= j; i--)
#define pb push_back
#define make make_pair
#define INF 0x3f3f3f3f
#define inf LLONG_MAX
#define PI acos(-1)
#define fir first
#define sec second
#define lb(x) ((x) & (-(x)))
#define dbg(x) cout<<#x<<" = "<<x<<endl;
using namespace std;

const int N = 1e6 + 5;

const LL mod = 998244353;

const ULL base = 101;

ULL pw_base[N], hs[N];

LL pw_ten[N], pre[N];

char a[N];

int id[N];

ULL get(int l, int r) { /// 获得区间 [l,r] 的哈希值

    return hs[r] - hs[l - 1] * pw_base[r - l + 1];

}

ULL get1(int st, int len, int pos) { /// 获得 x + A 字符串的前面 pos 个字符的哈希值

    if(pos <= len) return get(st + 1, st + pos);

    else return get(st + 1, st + len) * pw_base[pos - len] + get(1, pos - len);

}

bool cmp(int x, int y) {

    int mi = min(x, y); int ma = max(x, y); int dis = ma - mi;

    int l = 1, r = ma, ans = 0, mid = 0;

    while(l <= r) {

        mid = (l + r) >> 1;

        if(get1(mi, dis, mid) != hs[mid]) {

            ans = mid; r = mid - 1;

        }

        else l = mid + 1;

    }

    char ax, bx;

    ax = a[ans];

    if(ans <= dis) bx = a[mi + ans];

    else bx = a[ans - dis];

    if(ax > bx) return x < y;

    else return x >= y;

}

void solve() {

    scanf("%s", a + 1);

    int n = strlen(a + 1);

    pw_ten[0] = 1; pw_base[0] = 1;

    rep(i, 1, n) {

        id[i] = i;

        pw_ten[i] = pw_ten[i - 1] * 10 % mod;

        pre[i] = (pre[i - 1] * 10 % mod + (a[i] - '0')) % mod;

        pw_base[i] = pw_base[i - 1] * base;

        hs[i] = hs[i - 1] * base + a[i];

    }

    sort(id + 1, id + 1 + n, cmp);

    LL ans = 0LL;

    rep(i, 1, n) {

        ans = (ans * pw_ten[id[i]] % mod + pre[id[i]]) % mod;

    }

    printf("%lld
", ans);

}


int main() {

//    int _; scanf("%d", &_);
//    while(_--) solve();

    solve();

    return 0;
}