FFT与多项式、生成函数题目泛做

题目1 COGS 很强的乘法问题

高精度乘法用FFT加速

#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cmath>
#define Pi acos(-1)

using namespace std;
const int N = 600000 + 5;
const int rad = 10;

struct Complex {
  double r, i;
  Complex(double x = 0, double y = 0) { r = x; i = y; }
  Complex operator + (Complex x) { return Complex(r + x.r, i + x.i); }
  Complex operator - (Complex x) { return Complex(r - x.r, i - x.i); }
  Complex operator * (Complex x) { return Complex(r * x. r - i * x.i, r * x.i + i * x.r); }
}a[N], b[N];;

int n, m, l, r[N], c[N];
char ss[N];

void dft(Complex *s, int f) {
  for(int i = 0; i < n; ++ i)
    if(i < r[i]) swap(s[i], s[r[i]]);
  for(int i = 1; i < n; i <<= 1) {
    Complex wn(cos(Pi / i), f * sin(Pi / i));
    for(int j = 0; j < n; j += (i << 1)) {
      Complex w(1, 0);
      for(int k = 0; k < i; ++ k) {
        Complex x = s[j + k], y = w * s[j + k + i];
        s[j + k] = x + y;
        s[j + k + i] = x - y;
        w = w * wn;
      }
    }
  }
  if(f == -1)
    for(int i = 0; i < n; ++ i)
      s[i].r /= n;
}

int main() {
#ifndef stone
  freopen("bettermul.in", "r", stdin);
  freopen("bettermul.out", "w", stdout);
#endif

  int len1, len2;
  scanf("%s", ss); len1 = strlen(ss);
  for(int i = 0; i < len1; ++ i) a[i] = ss[len1 - i - 1] - '0';
  scanf("%s", ss); len2 = strlen(ss);
  for(int i = 0; i < len2; ++ i) b[i] = ss[len2 - i - 1] - '0';
  n = max(len1, len2); m = (n << 1) - 2;
  for(n = 1; n <= m; n <<= 1) l ++;
  for(int i = 0; i < n; ++ i)
    r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
  dft(a, 1); dft(b, 1);
  for(int i = 0; i < n; ++ i) a[i] = a[i] * b[i];
  dft(a, -1);
  for(int i = 0; i <= m; ++ i) c[i] = (int) (a[i].r + 0.5);
  for(int i = 0; i <= m; ++ i) {
    if(c[i] >= rad) {
      c[i + 1] += c[i] / rad;
      c[i] %= rad;
      if(i == m) m ++;
    }
  }
  int pos = m;
  while(c[pos] == 0) -- pos;
  for(int i = pos; i >= 0; -- i) printf("%d", c[i]);

#ifndef stone
  fclose(stdin); fclose(stdout);
#endif
  return 0;
}

题目2 BZOJ4259 残缺的字符串

算法讨论：（摘自Claris的博客，写得很清楚，我就是看这个才会做的）

注意：如果开第二个字符串的倍长的话，FFT会超时。还要注意统计答案的区间是什么。不要错了。

#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
 
using namespace std;
typedef long long ll;
const int N = 600000 + 5;
#define Pi acos(-1)
 
struct Cp {
  double r, i;
  Cp(double _r = 0, double _i = 0):
    r(_r), i(_i) {}
  Cp operator + (Cp x) {
    return Cp(r + x.r, i + x.i);
  }
  Cp operator - (Cp x) {
    return Cp(r - x.r, i - x.i);
  }
  Cp operator * (Cp x) {
    return Cp(r * x.r - i * x.i, r * x.i + i * x.r);
  }
};
 
char str1[N], str2[N];
int len1, len2, n, m, l, cnt;
int t[N], ans[N], r[N];
Cp a[N], b[N], A[N], B[N], C[N], tp;
 
 
void DFT(Cp *s, int f) {
  for(int i = 0; i < n; ++ i)
    if(i < r[i]) swap(s[i], s[r[i]]);
  for(int i = 1; i < n; i <<= 1) {
    Cp wn(cos(Pi / i), f * sin(Pi / i));
    for(int j = 0; j < n; j += (i << 1)) {
      Cp w(1, 0);
      for(int k = 0; k < i; ++ k) {
        Cp x = s[j + k], y = w * s[j + k + i];
        s[j + k] = x + y;
        s[j + k + i] = x - y;
        w = w * wn;
      }
    }
  }
  if(f == -1)
    for(int i = 0; i < n; ++ i)
      s[i].r /= n;
}
 
#define stone
 
int main() {
#ifndef stone
   
  freopen("bitch.in", "r", stdin);
  freopen("bitch.out", "w", stdout);
 
#endif
   
  scanf("%d%d", &len1, &len2);
  scanf("%s%s", str1, str2);
  reverse(str1, str1 + len1);
  for(int i = 0; i < len1; ++ i) {
    if(str1[i] == '*') a[i].r = 0;
    else a[i].r = (str1[i] - 'a' + 1);
  }
  for(int i = 0; i < len2; ++ i) {
    if(str2[i] == '*') b[i].r = 0;
    else b[i].r = (str2[i] - 'a' + 1);
  }
  m = len1 + len2;
  for(n = 1; n <= m; n <<= 1) ++ l;
  for(int i = 0; i < n; ++ i) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
 
  for(int i = 0; i < n; ++ i) {
    A[i].r = a[i].r * a[i].r * a[i].r;
    B[i].r = b[i].r;
  }
  DFT(A, 1); DFT(B, 1);
  for(int i = 0; i < n; ++ i) { C[i] = C[i] + A[i] * B[i]; }
  memset(A, 0, sizeof A); memset(B, 0, sizeof B);
  for(int i = 0; i < n; ++ i) {
    A[i].r = a[i].r * a[i].r;
    B[i].r = b[i].r * b[i].r;
  }
  DFT(A, 1); DFT(B, 1); tp = (Cp) {2, 0};
  for(int i = 0; i < n; ++ i) { C[i] = C[i] - A[i] * B[i] * tp; }
  memset(A, 0, sizeof A); memset(B, 0, sizeof B);
  for(int i = 0; i < n; ++ i) {
    A[i].r = a[i].r;
    B[i].r = b[i].r * b[i].r * b[i].r;
  }
  DFT(A, 1); DFT(B, 1);
  for(int i = 0; i < n; ++ i) {  C[i] = C[i] + A[i] * B[i]; }
  DFT(C, -1);
  for(int i = len1 - 1; i < len2; ++ i)
    if(C[i].r < 0.5) {
      ans[++ cnt] = i - len1 + 2;
    }
  printf("%d
", cnt);
  for(int i = 1; i <= cnt; ++ i) {
    printf("%d ", ans[i]);
  }
 
#ifndef stone
 
  fclose(stdin); fclose(stdout);
 
#endif
   
  return 0;
}

题目3 BZOJ 万径人踪灭

算法讨论：

用FFT算出所有回文串（相邻的和不相邻的），然后用Manacher计算出所有相邻的回文串，然后做差统计答案即可。

用FFT算出所有回文串的时候，是有len + len - 1个位置要计算的，分别是在字符的位置上和在两个字符之间的位置。

#include <cstdlib>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
 
using namespace std;
const int N = 400000 + 5;
typedef long long ll;
const int mod = 1e9 + 7;
#define Pi acos(-1)
 
struct Cp {
  double r, i;
  Cp(double _r = 0, double _i = 0) :
    r(_r), i(_i){}
  Cp operator + (Cp x) {
    return Cp(r + x.r, i + x.i);
  }
  Cp operator - (Cp x) {
    return Cp(r - x.r, i - x.i);
  }
  Cp operator * (Cp x) {
    return Cp(r * x.r - i * x.i, r * x.i + i * x.r);
  }
};
 
char ss[N], Ma[N << 1];
int n, m, l, r[N], lens;
Cp a[N], b[N];
ll Mp[N << 1], t1[N], t2[N], p[N];
 
void DFT(Cp *s, int f) {
  for(int i = 0; i < n; ++ i)
    if(i < r[i]) swap(s[i], s[r[i]]);
  for(int i = 1; i < n; i <<= 1) {
    Cp wn(cos(Pi / i), f * sin(Pi / i));
    for(int j = 0; j < n; j += (i << 1)) {
      Cp w(1, 0);
      for(int k = 0; k < i; ++ k) {
        Cp x = s[j + k], y = w * s[j + k + i];
        s[j + k] = x + y;
        s[j + k + i] = x - y;
        w = w * wn;
      }
    }
  }
  if(f == -1)
    for(int i = 0; i < n; ++ i)
      s[i].r /= n;
}
 
ll Manacher(char *s, int len) {
  int l = 0; ll res = 0;
  Ma[l ++] = '$';
  Ma[l ++] = '#';
  for(int i = 0; i < len; ++ i) {
    Ma[l ++] = s[i];
    Ma[l ++] = '#';
  }
  Ma[l] = '!';
  int Mx = 0, id = 0;
  for(int i = 0; i < l; ++ i) {
    if(Mx > i) {
      Mp[i] = min(Mp[2 * id - i], (ll)Mx - i);
    }
    else {
      Mp[i] = 1;
    }
    while(Ma[i + Mp[i]] == Ma[i - Mp[i]]) Mp[i] ++;
    if(i + Mp[i] > Mx) {
      Mx = i + Mp[i];
      id = i;
    }
    res += (Mp[i] >> 1); res %= mod;
  }
  return res;
}
 
int main() {
  ll ans = 0, tmp;
  scanf("%s", ss);
  n = strlen(ss); lens = strlen(ss);
  for(int i = 0; i < n; ++ i) {
    if(ss[i] == 'a')
      a[i].r = 1, b[i].r = 0;
    else a[i].r = 0, b[i].r = 1;
  }
  m = (n << 1) - 2;
  for(n = 1; n <= m; n <<= 1) l ++;
  for(int i = 0; i < n; ++ i) r[i] = (r[i >> 1] >> 1) |((i & 1) << (l - 1));
  DFT(a, 1); DFT(b, 1);
  for(int i = 0; i < n; ++ i) {
    a[i] = a[i] * a[i];
    b[i] = b[i] * b[i];
  }
  DFT(a, -1); DFT(b, -1);
  for(int i = 0; i <= n; ++ i) {
    t1[i] = (ll) (a[i].r + 0.5);
    t2[i] = (ll) (b[i].r + 0.5);
  }
  for(int i = 0; i < n; ++ i) {
    t1[i] += t2[i];
  }
  tmp = Manacher(ss, strlen(ss));
  p[0] = 1;
  for(int i = 1; i <= lens; ++ i)
    p[i] = p[i - 1] * 2 % mod;
  lens = lens + lens - 1;
  for(int i = 0; i < lens; ++ i) {
    if(i & 1) {
      ans = (ans + p[t1[i] >> 1] - 1) % mod;
    }
    else {
      ans = (ans + (p[(t1[i] - 1) >> 1] << 1) - 1) % mod;
    }
  }
  ans = (ans - tmp) % mod;
  while(ans < 0) {
    ans += mod;
    ans %= mod;
  }
  printf("%lld
", ans);
  return 0;
}

题目4 BZOJ3028 

首先很高兴的表示这里的汉堡是承德的。然后表示如果会生成函数的话这就是个输入输出题。

系数是种类，指数是数量性质。

如下图：

然后就没有然后了。知道答案就考虑输入和输出了。表示0ms过没有压力。这题如果你要是写个搜索。。。。。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstdlib>
#include <algorithm>
#include <cctype>
 
using namespace std;
const int N = 500 + 5;
const int mod = 10007;
typedef long long ll;
 
ll n;
 
inline ll read() {
  ll x = 0;
  char c = getchar();
  while(!isdigit(c)) c = getchar();
  while(isdigit(c)) {
    x = x * 10 + c - '0';
    x %= mod;
    c = getchar();
  }
  return x;
}
 
int main() {
  n = read();
  n = n * (n + 1) * (n + 2);
  n /= 6;
  n %= mod;
  printf("%lld
", n);
  return 0;
}