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求個以\(i\)為結尾的\(AA\)串的個數和以\(i\)為開頭的\(AA\)串的個數
乘法原理即可，暴力求有95分
而你會發現，枚舉l，經過\(i\)和\(i+l\)的只要算出它左右各能擴展到哪裏，然後這個區間內的都要\(+1\)
差分一下+後綴數組

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const
 
 int _(1e5 + 5);

IL ll Input(){
    RG ll x = 0, z = 1; RG char c = getchar();
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    return
 
 x * z;
}

int n, pre[_], suf[_], lg[_];
ll ans = 0;
char s[_];
struct SA{
    int a[_], t[_], tmp[_], rk[_], sa[_], height[_], st[20][_];

    IL void Init(){
        Fill(a, 0);
    }
    
    IL bool Cmp(RG int x, RG int y, RG int z){
        return tmp[x] == tmp[y] && x + z <= n && y + z <= n && tmp[x + z] == tmp[y + z];
    }

    IL void
 
 Suffix_Sort(){
        RG int m = 26;
        for(RG int i = 0; i <= m; ++i) t[i] = 0;
        for(RG int i = 1; i <= n; ++i) ++t[rk[i] = a[i]];
        for(RG int i = 1; i <= m; ++i) t[i] += t[i - 1];
        for(RG int i = n; i; --i) sa[t[rk[i]]--] = i;
        for(RG int k = 1; k <= n; k <<= 1){
            RG int l = 0;
            for(RG int i = n - k + 1; i <= n; ++i) tmp[++l] = i;
            for(RG int i = 1; i <= n; ++i) if(sa[i] > k) tmp[++l] = sa[i] - k;
            for(RG int i = 0; i <= m; ++i) t[i] = 0;
            for(RG int i = 1; i <= n; ++i) ++t[rk[tmp[i]]];
            for(RG int i = 1; i <= m; ++i) t[i] += t[i - 1];
            for(RG int i = n; i; --i) sa[t[rk[tmp[i]]]--] = tmp[i];
            swap(rk, tmp); rk[sa[1]] = l = 1;
            for(RG int i = 2; i <= n; ++i) rk[sa[i]] = Cmp(sa[i - 1], sa[i], k) ? l : ++l;
            if(l >= n) break;
            m = l;
        }
        for(RG int i = 1, h = 0; i <= n; ++i){
            if(h) --h;
            while(a[i + h] == a[sa[rk[i] - 1] + h]) ++h;
            height[rk[i]] = h;
        }
    }

    IL void ST_Prepare(){
        for(RG int i = 1; i <= n; ++i) st[0][i] = height[i];
        for(RG int i = 1; i <= lg[n]; ++i)
            for(RG int j = 1; j + (1 << i) - 1 <= n; ++j)
                st[i][j] = min(st[i - 1][j], st[i - 1][j + (1 << (i - 1))]);
    }

    IL int LCP(RG int x, RG int y){
        x = rk[x]; y = rk[y];
        if(x > y) swap(x, y);
        RG int len = lg[y - x];
        return min(st[len][x + 1], st[len][y - (1 << len) + 1]);
    }
} A, B;

int main(RG int argc, RG char* argv[]){
    for(RG int i = 2; i <= 30000; ++i) lg[i] = lg[i >> 1] + 1;
    for(RG int T = Input(); T; --T){
        A.Init(); B.Init();
        scanf(" %s", s + 1);
        n = strlen(s + 1); ans = 0;
        for(RG int i = 1; i <= n; ++i){
            B.a[n - i + 1] = A.a[i] = s[i] - 'a' + 1;
            suf[i] = pre[i] = 0;
        }
        A.Suffix_Sort(); A.ST_Prepare();
        B.Suffix_Sort(); B.ST_Prepare();
        for(RG int l = 1; l + l <= n; ++l)
            for(RG int i = l; i + l <= n; i += l){
                RG int r = i + l, x = min(A.LCP(i, r), l), y = min(B.LCP(n - i + 1, n - r + 1), l), len = x + y - l;
                if(x + y > l){
                    suf[i - y + 1]++; suf[i - y + 1 + len]--;
                    pre[r + x - len]++; pre[r + x]--;
                }
            }
        for(RG int i = 1; i <= n; ++i) pre[i] += pre[i - 1], suf[i] += suf[i - 1];
        for(RG int i = 1; i < n; ++i) ans += 1LL * pre[i] * suf[i + 1];
        printf("%lld\n", ans);
    }
    return 0;
}

BZOJ4650 : [NOI2016]優秀的拆分