1 条题解

  • 0
    @ 2023-6-21 20:06:08

    C++ :

    #pragma GCC optimize ("O2")
    #define ONLINE_JUDGE
    #include<bits/stdc++.h>
    
    #define HEAP priority_queue
    #define rep(i, n) for(int i = 0, _end_ = (n); i < _end_; ++i)
    #define per(i, n) for(int i = (n) - 1; i >= 0 ; --i)
    #define forn(i, l, r) for(int i = (l), _end_ = (r); i <= _end_; ++i)
    #define nrof(i, r, l) for(int i = (r), _end_ = (l); i >= _end_; --i)
    #define FOR(a, b) for(auto (a): (b))
    #define mp make_pair
    #define mt make_tuple
    #define pb push_back
    #define X first
    #define Y second
    #define eps 1e-6
    #define pi 3.1415926535897932384626433832795
    #define SZ(x) (int)x.size()
    #define ALL(x) x.begin(), x.end()
    #define FILL(a, b) memset((a), (b), sizeof((a)))
    #define MCPY(a, b) memcpy((a), (b), sizeof((b)))
    
    using namespace std;
    
    typedef long long LL;
    typedef double flt;
    typedef vector<int> vi;
    typedef vector<LL> vl;
    typedef pair<int,int> pii;
    typedef pair<int,LL> pil;
    typedef pair<LL,int> pli;
    typedef pair<LL,LL> pll;
    typedef vector<pil> vil;
    typedef vector<pii> vii;
    typedef vector<pli> vli;
    typedef vector<pll> vll;
    
    const int iinf = 1e9 + 7;
    const int oo = 0x3f3f3f3f;
    const LL linf = 1ll << 60;
    const flt dinf = 1e60;
    
    template <typename T>
    inline void scf(T &x)
    {
    	bool f = 0; x = 0; char c = getchar();
    	while((c < '0' || c > '9') && c != '-') c = getchar();
    	if(c == '-') { f = 1; c = getchar(); }
    	while(c >= '0' && c <= '9') { x = x * 10 + c - '0'; c = getchar(); }
    	if(f) x = -x; return;
    }
    template <typename T1, typename T2>
    void scf(T1 &x, T2 &y) { scf(x); return scf(y); }
    template <typename T1, typename T2, typename T3>
    void scf(T1 &x, T2 &y, T3 &z) { scf(x); scf(y); return scf(z); }
    template <typename T1, typename T2, typename T3, typename T4>
    void scf(T1 &x, T2 &y, T3 &z, T4 &w) { scf(x); scf(y); scf(z); return scf(w); }
    
    inline char mygetchar(){ char c = getchar(); while(c == ' ' || c == '\n') c = getchar(); return c; }
    
    template <typename T> inline bool chkmax(T &x, const T &y){ return y > x ? x = y, 1 : 0; }
    template <typename T> inline bool chkmin(T &x, const T &y){ return y < x ? x = y, 1 : 0; }
    
    #ifdef ONLINE_JUDGE
    #define debug(...) ;
    #else
    #define debug(...) fprintf(stderr, __VA_ARGS__)
    #define DEBUG
    #endif
    
    //---------------------------------------------------------head----------------------------------------------------
    
    const int maxn = 1e5 + 100;
    
    const int maxb = 10;
    
    const int mod = 1e9 + 7;
    
    inline void ADD(int &x, const int &y){ x = (x + y < mod) ? (x +y) : (x + y - mod); }
    inline int SUM(const int &x, const int &y){ return (x + y < mod) ? (x +y) : (x + y - mod); }
    inline void SUB(int &x, const int &y){ x = (x - y < 0) ? (x - y + mod) : (x - y); }
    inline int DIF(const int &x, const int &y){ return (x - y < 0) ? (x - y + mod) : (x - y); }
    
    int n, q;
    char str[maxn];
    int L[maxn], R[maxn], ans[maxn], ord[maxn], nxt[maxn][maxb], prv[maxn][maxb];
    
    void work(int ln, int rn, int lq, int rq)
    {
    	if(ln == rn)
    	{
    		forn(i, lq, rq) ans[ord[i]] = 1;
    		return;
    	}
    	if(lq > rq) return;
    
    	int midn = ln + rn >> 1;
    	static int sum[10], rem[10], f[maxn][10], dp[maxn], F[maxn][10], G[maxn][10], DP[maxn], sm, s[10][10];
    	memset(sum, 0, sizeof sum); memset(rem, 0, sizeof rem);
    	nrof(i, midn, ln)
    	{
    		rep(j, maxb) f[i][j] = SUM((nxt[i][j] <= midn) ? 0 : 1, sum[j]);
    		if(rem[str[i] - 'a']) rep(j, maxb) SUB(sum[j], f[rem[str[i] - 'a']][j]);
    		rep(j, maxb) ADD(sum[j], f[i][j]);
    		rem[str[i] - 'a'] = i;
    	}
    	memset(sum, 0, sizeof sum); memset(rem, 0, sizeof rem);
    	nrof(i, midn, ln)
    	{
    		if(rem[str[i] - 'a']) rep(j, maxb) SUB(sum[j], f[rem[str[i] - 'a']][j]);
    		rep(j, maxb)
    		{
    			ADD(sum[j], f[i][j]), F[i][j] = sum[j];
    			if(nxt[i - 1][j] > midn) ADD(F[i][j], 1);
    		}
    		rem[str[i] - 'a'] = i;
    	}
    	memset(s, 0, sizeof s); memset(sum, 0, sizeof sum);
    	forn(i, midn + 1, rn)
    	{
    		rep(j, maxb) G[i][j] = SUM((str[i] - 'a' == j && prv[i][j] <= midn) ? 1 : 0, DIF(sum[j], s[str[i] - 'a'][j]));
    		memcpy(s[str[i] - 'a'], sum, sizeof sum);
    		rep(j, maxb) ADD(sum[j], G[i][j]);
    	}
    	forn(i, midn + 2, rn) rep(j, maxb) ADD(G[i][j], G[i - 1][j]);
    	sm = 0; memset(rem, 0, sizeof rem);
    	nrof(i, midn, ln)
    	{
    		dp[i] = SUM(sm, 1);
    		if(rem[str[i] - 'a']) SUB(sm, dp[rem[str[i] - 'a']]);
    		ADD(sm, dp[i]);
    		rem[str[i] - 'a'] = i;
    	}
    	sm = 0; memset(rem, 0, sizeof rem);
    	nrof(i, midn, ln)
    	{
    		if(rem[str[i] - 'a']) SUB(sm, dp[rem[str[i] - 'a']]);
    		ADD(sm, dp[i]), DP[i] = sm;
    		rem[str[i] - 'a'] = i;
    	}
    
    	static int qa[maxn], qb[maxn], na, nb;
    	na = nb = 0;
    	forn(_, lq, rq)
    	{
    		int &i = ord[_];
    		if(L[i] > midn) qb[nb++] = i;
    		else if(R[i] <= midn) qa[na++] = i;
    		else
    		{
    			ans[i] = DP[L[i]];
    			rep(j, maxb) ADD(ans[i], 1ll * F[L[i]][j] * G[R[i]][j] % mod);
    		}
    	}
    	int foo, bar;
    	rep(i, na) ord[lq + i] = qa[i];
    	foo = lq + na;
    	rep(i, nb) ord[foo + i] = qb[i];
    	bar = foo + nb;
    	work(ln, midn, lq, foo - 1);
    	work(midn + 1, rn, foo, bar - 1);
    	return;
    }
    
    int main()
    {
    	scanf("%s", str + 1); n = strlen(str + 1);
    	scf(q); forn(i, 1, q) scf(L[i], R[i]), ord[i] = i;
    	rep(i, maxb) nxt[n][i] = n;
    	nrof(i, n - 1, 0){ memcpy(nxt[i], nxt[i + 1], sizeof nxt[i]); nxt[i][str[i + 1] - 'a'] = i + 1; }
    	rep(i, maxb) prv[1][i] = 1;
    	forn(i, 2, n){ memcpy(prv[i], prv[i - 1], sizeof prv[i]); prv[i][str[i - 1] - 'a'] = i - 1; }
    	work(1, n, 1, q);
    	forn(i, 1, q) printf("%d\n", ans[i]);
    	return 0;
    }
    
    • 1

    #6074. 「2017 山东一轮集训 Day6」子序列

    信息

    ID
    1030
    时间
    2000ms
    内存
    512MiB
    难度
    (无)
    标签
    递交数
    0
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