11.10 解题报告

T1

考场用时：\(30\) min
期望得分：\(100\) pts
实际得分：\(100\) pts
直接定义一个分数类，重载 + 和 *，然后拓扑即可。

#include<bits/stdc++.h>
#define ll __int128 
#define int __int128
#define lc(k) k<<1
#define rc(k) k<<1|1
#define lb lower_bound
#define orz puts("gzn ak ioi")
const int MAX=1e5+10;
const int MOD=1e9+7;
using namespace std;
inline char readchar() {
	static char buf[100000], *p1 = buf, *p2 = buf;
	return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF : *p1++;
}
inline int read() {
	int res = 0, f = 0;
	char ch = readchar();
	for(; !isdigit(ch); ch = readchar()) if(ch == '-') f = 1;
	for(; isdigit(ch);ch = readchar()) res = (res << 1) + (res << 3) + (ch ^ '0');
	return f ? -res : res;
}
inline void write(int x) {
    if(x<0){putchar('-');x=-x;}
    if(x>9) write(x/10);
    putchar(x%10+'0');
}
int n,m,cd[MAX],rd[MAX];// 0 分子 1 分母 
int head[MAX],cnt;
struct node{int net,v;}e[MAX*100];
void add(int u,int v){
	e[++cnt]=(node){head[u],v};
	head[u]=cnt;
	return ;
}
struct fs{
	int fz,fm;
	friend fs operator + (fs x,fs y){
		fs ret;
		ret.fz=x.fm*y.fz+x.fz*y.fm;
		ret.fm=x.fm*y.fm;
		int gcd=__gcd(ret.fz,ret.fm);
		ret.fz/=gcd;ret.fm/=gcd;
		return ret;
	}
	friend fs operator * (fs x,fs y){
		fs ret;
		ret.fz=x.fz*y.fz;
		ret.fm=x.fm*y.fm;
		int gcd=__gcd(ret.fz,ret.fm);
		ret.fz/=gcd;ret.fm/=gcd;
		return ret;
	}
}ans[MAX];
queue<int> q;
void topo(){
	for(int i=1;i<=n;i++){
		ans[i].fm=1;
		if(!rd[i]){
			ans[i].fz=1;
			q.push(i);
		}
	}
	while(!q.empty()){
		int u=q.front();q.pop();
//		write(u);puts("");
		fs t=(fs){1,cd[u]};
		for(int i=head[u];i;i=e[i].net){
			int v=e[i].v;
			ans[v]=ans[v]+ans[u]*t;
			rd[v]--;
			if(!rd[v]) q.push(v);
		}
	}
	return ;
}
signed main(){
//	freopen("water3.in","r",stdin);
//	freopen("1.out","w",stdout);
	n=read(),m=read();
	for(int i=1;i<=n;i++){
		int x=read();
		for(int j=1;j<=x;j++){
			int v=read();
			add(i,v);
			cd[i]++;rd[v]++;
		}
	}
	topo();
	for(int i=1;i<=n;i++)
		if(!cd[i]){
			write(ans[i].fz);
			putchar(' ');
			write(ans[i].fm);
			putchar('\n');
		}
	return 0;
}

T2

考场用时：\(1\) h
期望得分：\(55\) pts
实际得分：\(55\) pts
发现 第一个 A /唯一的 C 必然分别是一个前缀 / 后缀
预处理 每个位置的 后缀奇数个字符个数（与 C 有关）
每个位置的 前缀奇数个数分别为 i 的位置个数 （与 A 有关），并统计其前缀和
哈希来寻找匹配 AB 这一循环节

#include<bits/stdc++.h>
#define LL long long
#define lc(k) k<<1
#define rc(k) k<<1|1
#define orz cout<<"I AK IOI\n"
//#define int long long
const int MAX=2e5+10;
const int MOD=998244353;
using namespace std;
inline char readchar() {
	static char buf[100000], *p1 = buf, *p2 = buf;
	return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF : *p1++;
}
inline int read() {
#define readchar getchar
	int res = 0, f = 0;
	char ch = readchar();
	for(; !isdigit(ch); ch = readchar()) if(ch == '-') f = 1;
	for(; isdigit(ch); ch = readchar()) res = (res << 1) + (res << 3) + (ch ^ '0');
	return f ? -res : res;
}
inline void write(int x) {
    if(x<0){putchar('-');x=-x;}
    if(x>9) write(x/10);
    putchar(x%10+'0');
}
const int Maxc = 30;
const int Maxn = (1 << 20) + 5;
const LL Mod1 = 998244353; 
const LL Mod2 = 19260817;

LL hash1[Maxn], hash2[Maxn], pow26[Maxn], Pow26[Maxn];//前缀哈希值

inline void init0() {
	pow26[0] = 1;
	Pow26[0] = 1; 
	for(register int i = 1; i < Maxn; ++i) {
		pow26[i] = pow26[i - 1] * 26; pow26[i] %= Mod1;
		Pow26[i] = Pow26[i - 1] * 26; Pow26[i] %= Mod2;
	}
}

inline LL gethash1(int l, int r) {
	return ((hash1[r] % Mod1 - 1LL * hash1[l - 1] * pow26[r - l + 1] % Mod1) % Mod1 + Mod1) % Mod1;
}
inline LL gethash2(int l, int r) {
	return ((hash2[r] % Mod2 - 1LL * hash2[l - 1] * Pow26[r - l + 1] % Mod2) % Mod2 + Mod2) % Mod2;
}
bool check(int lf, int rt, int l, int r) {
	return (gethash1(lf, rt) == gethash1(l, r)) && (gethash2(lf, rt) == gethash2(l, r));
}

int s[Maxn]; // s[i][j]:前 i 个位置有多少个位置奇数个字符不大于 j 
int S[Maxn][Maxc];
int bs[Maxn];
int cnt[Maxc];
string str;
int T, ans;

int n, a[Maxn];

void Init() {
	ans = 0;
	memset(cnt, 0, sizeof(cnt));
	memset(bs, 0, sizeof(bs));
	memset(s, 0, sizeof(s));
	memset(hash1, 0, sizeof(hash1));
	memset(hash2, 0, sizeof(hash2));
	for(register int i = 1; i <= n; ++i) {
		hash1[i] = hash1[i - 1] * 26 % Mod1 + a[i];
		hash1[i] %= Mod1;
		hash2[i] = hash2[i - 1] * 26 % Mod2 + a[i];
		hash2[i] %= Mod2;
	}
	for(int i = 1; i <= n; ++i) {
		cnt[a[i]]++;
		if(cnt[a[i]] & 1) s[i] = s[i - 1] + 1;
		else s[i] = s[i - 1] - 1;
		for(int j = 0; j <= 26; ++j) {
			S[i][j] = S[i - 1][j];
			if(j >= s[i]) S[i][j]++;
		}
	}
	memset(cnt, 0, sizeof(cnt));
	for(int i = n; i >= 1; --i) {
		cnt[a[i]]++;
		if(cnt[a[i]] & 1) bs[i] = bs[i + 1] + 1;
		else bs[i] = bs[i + 1] - 1;
	}
	return;
}

void mainfunction() {
	cin >> str; n = str.size();
	for(int i = 0; i < str.size(); ++i) a[i + 1] = (int)str[i] - (int)'a';
	Init();
	for(int i = 2; i < n; ++i) {
		bool flag = 1;
		for(int j = i, lt = 1; (j < n) && flag; j += i, lt += i) {
			if(check(1, i, lt, j)) {
				ans += S[i - 1][bs[j + 1]];
			}
			else break; 
		}
	}
	printf("%d\n", ans);
	return;
}

int main() {
	init0(); 
	T=read();
	while(T--) mainfunction();
	return 0;
}