NOIP2020T2题解

【题目大意】
题目是求 S = (AB)iC 的方案数,其中 F(A) ≤ F©,F(S ) 表示字符串 S中出现奇数次的字符的数量。
【解题思路】
先确定好C的可能,即末尾几位字符。然后再考虑划分(AB)i,需要找出(AB)i的最小重复子串,相关算法考虑KMP算法或者字符串哈希,再在该子串中划分A和B,产生方案。
需要注意的是AB不一定只能存在于最小的重复子串中。比如:
(AB)i = abababababab,最小重复是 ab,但重复的部分也可以是 abab,同样可以在abab中划分AB串,另外还有ababab以及整个(AB)i。分别是最小重复串的1、2、3、6倍,都是最大重复次数6的约数。
另外定好(AB)的范围后,划分AB时注意题目限制F(A) ≤ F©。
【参考代码】

#include <bits/stdc++.h>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
const ll N = 1100000, base1 = 1021831, base2 = 96269; 
ull hash1[N], hash2[N], pow1[N], pow2[N];
ull f1(ll l, ll r) {
    
    
	return hash1[r] - hash1[l - 1] * pow1[r - l + 1];
}
ull f2(ll l, ll r) {
    
    
	return hash2[r] - hash2[l - 1] * pow2[r - l + 1];
}
pair<ull, ull> f(ll l, ll r) {
    
    
	return make_pair(f1(l,r), f2(l,r));
}
struct BIT {
    
    
    ll sum[30], n;
    void clear() {
    
    
		memset(sum, 0, sizeof(sum));
	}
    inline ll lowbit(ll x) {
    
    
		return x & (-x);
	}
    void add(ll x, ll v) {
    
    
		while (x <= n) {
    
    
			sum[x] += v; x += lowbit(x);
		}
	}
    ll find(ll x) {
    
    
		ll ans = 0;
		while(x) {
    
    
			ans += sum[x]; x -= lowbit(x);
		}
		return ans;
	}
} T;
ll book[30], tot[N], num[N];
char ch[N];
int main() {
    
    
	ll ans = 0, cnt = 0;
    ll Q; cin >> Q; T.n = 26;
    while (Q--) {
    
    
        T.clear();
        memset(book, 0, sizeof(book)); memset(tot, 0, sizeof(tot)); memset(num, 0, sizeof(num));
        cnt = ans = 0;
    	cin >> ch + 1; int n = strlen(ch + 1);
        hash1[0] = hash2[0] = 0; pow1[0] = pow2[0] = 1;
        for (int i = 1; i <= n; ++i) {
    
    
            hash1[i] = hash1[i - 1] * base1 + ch[i]; hash2[i] = hash2[i - 1] * base2 + ch[i];
            pow1[i] = pow1[i - 1] * base1; pow2[i] = pow2[i - 1] * base2;
        }
        for (int i = n; i >= 3; --i) {
    
    
            book[ch[i] - 'a'] ^= 1;
            if (!book[ch[i] - 'a']) {
    
    
				tot[i] = --cnt;
			} else {
    
    
				tot[i] = ++cnt;
			}
        }
        num[n - 1] = 1;
        for (int i = n - 2; i >= 2; --i) {
    
    
        	if (i * 2 > n - 1 || f(1, i) != f(i + 1, i * 2)) {
    
    
				num[i] = 1;
				continue;
			}
        	num[i] = num[i << 1] << 1;
        	if (num[i] * i + i <= n - 1 && f(1, i) == f(num[i] * i + 1, num[i] * i + i)) {
    
    
				++num[i];
			}
		}
		cnt = 0; memset(book, 0, sizeof(book));
        for (int i = 2; i < n; ++i) {
    
    
            book[ch[i - 1] - 'a'] ^= 1;
            if (!book[ch[i - 1] - 'a']) {
    
    
				--cnt;
			} else {
    
    
				++cnt;
			} 
            T.add(cnt + 1, 1);
            ans += T.find(tot[i + 1] + 1) * ((num[i] >> 1) + (num[i] & 1)) + T.find(tot[i * 2 + 1] + 1) * (num[i] >> 1);
        }
        cout << ans << '\n';
    }
    return 0;
}

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转载自blog.csdn.net/yueyuedog/article/details/112120497