目录

noip模拟47

T1 prime

水题,只需要把1到 $min(k, \sqrt{R})$ 的质数都先筛出来,然后标记上在 $[max(2 * prime, L),R]$之间的质数就行了(然而我还是挂分了)

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#include <bits/stdc++.h>
#define For(i, l, r) for (long long i = (l), i##end = (r); i <= i##end; ++i)
#define Fordown(i, r, l) for (long long i = (r), i##end = (l); i >= i##end; --i)
#define Set(a, v) memset(a, v, sizeof(a))
using namespace std;
const long long maxn = 10000010;

template<typename LDXIN>
LDXIN read() {
  LDXIN s = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    s = s * 10 + ch - '0';
    ch = getchar();
  }
  return s * f;
}

long long prime[maxn], tot;
long long l, r, k;
bool vis[maxn];

void s1(long long l, long long r) {
  long long ans = 0;
  For(i, l, r) {
    ans ^= i;
  }
  cout << ans << endl;
}

bool viS[maxn];
int main() {
  l = read<long long>();
  r = read<long long>();
  k = read<long long>();
  k = min(k, (long long)sqrt(r));
  vis[1] = 1;
  if (k == 1) {
    s1(l, r);
    return 0;
  }
  for (long long i = 2; i <= k;i++) {
    if (!vis[i])
      prime[++tot] = i;
    for (long long j = 1; j <= tot && prime[j] <= k && i * prime[j] <= k; j++) {
      vis[i * prime[j]] = 1;
      if (i % prime[j] == 0) break;
    }
  }
  for (int i = 1; i <= tot; i++) {
    long long stt = max(2 * prime[i], l) / prime[i];
    if (stt * prime[i] < l) stt++;
    while (prime[i] * stt <= r) {
      viS[prime[i] * stt - l] = 1;
      stt++;
    }
  }
  long long ans = 0;
  for (long long i = l; i <= r; i++) {
    
    if (viS[i - l] != 1) {
      ans ^= i;
    }
  }
  cout << ans << endl;
  return 0;
}

T2 Sequence

绝了!最后一个点狂T不止,但我认为思路莫得问题

/image/noip47/1.png

先写出最基础的DP程序,用 $dp[i]$ 表示以i结尾的字串的个数。每新增一个字母,它的dp值就等于 $1 + \Sigma^{k}_{i=1}dp[i]$ 可以使用前缀和优化,可以得到67分的高分

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#include <bits/stdc++.h>
#define For(i, l, r) for (int i = (l), i##end = (r); i <= i##end; ++i)
#define Fordown(i, r, l) for (int i = (r), i##end = (l); i >= i##end; --i)
#define Set(a, v) memset(a, v, sizeof(a))
using namespace std;
const long long maxn = 1000100;
const long long mod = 1e9 + 7;

template<typename LDXIN>
LDXIN read() {
  LDXIN s = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    s = s * 10 + ch - '0';
    ch = getchar();
  }
  return s * f;
}

long long n, m, k;
long long seq[maxn], dp[maxn];

struct PNT {
  long long id, val;
  friend bool operator< (PNT a, PNT b) {
    return a.val > b.val;
  }
};

int main() {
  n = read<long long>();
  m = read<long long>();
  k = read<long long>();
  For(i, 1, n) {
    seq[i] = read<long long>();
  }
  dp[seq[1]] = 1;
  long long sum = 1, tmp = 0;
  For(i, 2, n) {
    tmp = dp[seq[i]] % mod;
    
    dp[seq[i]] = sum + 1;
    dp[seq[i]] %= mod;
    
    sum = (dp[seq[i]] % mod - tmp % mod + mod + sum) % mod;
    sum %= mod;
  }
  priority_queue<PNT> pq;
  For(i, 1, k) {
    pq.push((PNT){i, dp[i]});
  }
  For(i, n + 1, m + n) {
    long long minVal = pq.top().val, minid = pq.top().id;
    pq.pop();
    seq[i] = minid;
    
    tmp = minVal;
    tmp %= mod;
    
    dp[seq[i]] = sum + 1;
    // cout << dp[seq[i]] << endl;
    dp[seq[i]] %= mod;
    
    sum = (dp[seq[i]] % mod - tmp % mod + mod + sum) % mod;    
    sum %= mod;
    
    pq.push((PNT){minid, dp[seq[i]]});
  }
  cout << sum % mod << endl;
  return 0;
}

然后你推了推几组数据,发现无论接下来的元素是什么,新的dp[x]都是一样的。于是可以贪心地填当前的dp值最小的那个元素,其实也就是最后出现位置最靠前的那个元素

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#include <bits/stdc++.h>
#define For(i, l, r) for (int i = (l), i##end = (r); i <= i##end; ++i)
#define Fordown(i, r, l) for (int i = (r), i##end = (l); i >= i##end; --i)
#define Set(a, v) memset(a, v, sizeof(a))
using namespace std;
const long long maxn = 10001100;
const long long mod = 1e9 + 7;

template<typename LDXIN>
LDXIN read() {
  LDXIN s = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    s = s * 10 + ch - '0';
    ch = getchar();
  }
  return s * f;
}

// #define DEBUG 1
struct IO {
#define MAXSIZE (1 << 20)
#define isdigit(x) (x >= '0' && x <= '9')
  char buf[MAXSIZE], *p1, *p2;
  char pbuf[MAXSIZE], *pp;
#if DEBUG
#else
  IO() : p1(buf), p2(buf), pp(pbuf) {}
  ~IO() { fwrite(pbuf, 1, pp - pbuf, stdout); }
#endif
  inline char gc() {
#if DEBUG
    return getchar();
#endif
    if (p1 == p2) p2 = (p1 = buf) + fread(buf, 1, MAXSIZE, stdin);
    return p1 == p2 ? ' ' : *p1++;
  }
  inline bool blank(char ch) {
    return ch == ' ' || ch == '\n' || ch == '\r' || ch == '\t';
  }
  template <class T>
  inline void read(T &x) {
    register double tmp = 1;
    register bool sign = 0;
    x = 0;
    register char ch = gc();
    for (; !isdigit(ch); ch = gc())
      if (ch == '-') sign = 1;
    for (; isdigit(ch); ch = gc()) x = x * 10 + (ch - '0');
    if (ch == '.')
      for (ch = gc(); isdigit(ch); ch = gc())
        tmp /= 10.0, x += tmp * (ch - '0');
    if (sign) x = -x;
  }
  inline void read(char *s) {
    register char ch = gc();
    for (; blank(ch); ch = gc())
      ;
    for (; !blank(ch); ch = gc()) *s++ = ch;
    *s = 0;
  }
  inline void read(char &c) {
    for (c = gc(); blank(c); c = gc())
      ;
  }
  inline void push(const char &c) {
#if DEBUG
    putchar(c);
#else
    if (pp - pbuf == MAXSIZE) fwrite(pbuf, 1, MAXSIZE, stdout), pp = pbuf;
    *pp++ = c;
#endif
  }
  template <class T>
  inline void write(T x) {
    if (x < 0) x = -x, push('-');
    static T sta[35];
    T top = 0;
    do {
      sta[top++] = x % 10, x /= 10;
    } while (x);
    while (top) push(sta[--top] + '0');
  }
  template <class T>
  inline void write(T x, char lastChar) {
    write(x), push(lastChar);
  }
} io;



long long n, m, k;
int seq[maxn];
struct DP {
  int val, lst = 0;
  friend bool operator< (DP a, DP b) {
    return a.lst < b.lst;
  }
} dp[maxn];

class MTX {
public:
  struct matrix {
    int a[105][105];
  } ans, base;
  int m;
  void init(int k) {
    m = k + 1;
    For(i, 1, k) {
      ans.a[i][1] = dp[i].val;
    }
    ans.a[k + 1][1] = 1;
    For(i, 1, k - 1) {
      base.a[i][i + 1] = 1;
    }
    For(i, 1, k + 1) {
      base.a[k][i] = 1;
    }
    base.a[k + 1][k + 1] = 1;
  }
  matrix calc(matrix a, matrix b) {
    matrix c;
    for (int i = 1; i <= m; i++) {
      for (int j = 1; j <= m; j++) {
        c.a[i][j] = 0;
      }
    }
    for (int i = 1; i <= m; i++) {
      for (int j = 1; j <= m; j++) {
        for (int k = 1; k <= m; k++) {
          c.a[i][j] = (c.a[i][j] + 1ll * b.a[i][k] * a.a[k][j]) % mod;
        }
      }
    }
    return c;
  }
  
  void qpow(int power) {
    while (power) {
      if (power & 1) {
        ans = calc(ans, base);
      }
      base = calc(base, base);
      power >>= 1;
    }
  }
} wdnmd;


int32_t main() {
  io.read<long long>(n);
  io.read<long long>(m);
  io.read<long long>(k);
  For(i, 1, n) {
    io.read<int>(seq[i]);
    dp[seq[i]].lst = i;
  }
  dp[seq[1]].val = 1;
  long long sum = 1, tmp = 0;
  For(i, 2, n) {
    tmp = dp[seq[i]].val;
    dp[seq[i]].val = sum + 1;
    dp[seq[i]].val %= mod;
    sum = (dp[seq[i]].val - tmp + sum + mod) % mod;
  }
  sum = 0;
  sort(dp + 1, dp + 1 + k);
  wdnmd.init(k);

  wdnmd.qpow(m);
  For(i, 1, k) {
    sum += wdnmd.ans.a[i][1];
  }
  cout << sum % mod << endl;
  return 0;
}