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 =  + ;
 const int rad = ;

 struct Complex {
   double r, i;
   Complex(double x = , double y = ) { 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 = ; i < n; ++ i)
     if(i < r[i]) swap(s[i], s[r[i]]);
   for(int i = ; i < n; i <<= ) {
     Complex wn(cos(Pi / i), f * sin(Pi / i));
     for(int j = ; j < n; j += (i << )) {
       Complex w(, );
       for(int k = ; 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 == -)
     for(int i = ; 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 = ; i < len1; ++ i) a[i] = ss[len1 - i - ] - '';
   scanf("%s", ss); len2 = strlen(ss);
   for(int i = ; i < len2; ++ i) b[i] = ss[len2 - i - ] - '';
   n = max(len1, len2); m = (n << ) - ;
   for(n = ; n <= m; n <<= ) l ++;
   for(int i = ; i < n; ++ i)
     r[i] = (r[i >> ] >> ) | ((i & ) << (l - ));
   dft(a, ); dft(b, );
   for(int i = ; i < n; ++ i) a[i] = a[i] * b[i];
   dft(a, -);
   for(int i = ; i <= m; ++ i) c[i] = (int) (a[i].r + 0.5);
   for(int i = ; i <= m; ++ i) {
     if(c[i] >= rad) {
       c[i + ] += c[i] / rad;
       c[i] %= rad;
       if(i == m) m ++;
     }
   }
   int pos = m;
   while(c[pos] == ) -- pos;
   for(int i = pos; i >= ; -- i) printf("%d", c[i]);

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

COGS

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 =  + ;
 #define Pi acos(-1)

 struct Cp {
   double r, i;
   Cp(double _r = , double _i = ):
     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 = ; i < n; ++ i)
     if(i < r[i]) swap(s[i], s[r[i]]);
   for(int i = ; i < n; i <<= ) {
     Cp wn(cos(Pi / i), f * sin(Pi / i));
     for(int j = ; j < n; j += (i << )) {
       Cp w(, );
       for(int k = ; 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 == -)
     for(int i = ; 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 = ; i < len1; ++ i) {
     if(str1[i] == '*') a[i].r = ;
     else a[i].r = (str1[i] - 'a' + );
   }
   for(int i = ; i < len2; ++ i) {
     if(str2[i] == '*') b[i].r = ;
     else b[i].r = (str2[i] - 'a' + );
   }
   m = len1 + len2;
   for(n = ; n <= m; n <<= ) ++ l;
   for(int i = ; i < n; ++ i) r[i] = (r[i >> ] >> ) | ((i & ) << (l - ));

   for(int i = ; i < n; ++ i) {
     A[i].r = a[i].r * a[i].r * a[i].r;
     B[i].r = b[i].r;
   }
   DFT(A, ); DFT(B, );
   for(int i = ; i < n; ++ i) { C[i] = C[i] + A[i] * B[i]; }
   memset(A, , sizeof A); memset(B, , sizeof B);
   for(int i = ; i < n; ++ i) {
     A[i].r = a[i].r * a[i].r;
     B[i].r = b[i].r * b[i].r;
   }
   DFT(A, ); DFT(B, ); tp = (Cp) {, };
   for(int i = ; i < n; ++ i) { C[i] = C[i] - A[i] * B[i] * tp; }
   memset(A, , sizeof A); memset(B, , sizeof B);
   for(int i = ; i < n; ++ i) {
     A[i].r = a[i].r;
     B[i].r = b[i].r * b[i].r * b[i].r;
   }
   DFT(A, ); DFT(B, );
   for(int i = ; i < n; ++ i) {  C[i] = C[i] + A[i] * B[i]; }
   DFT(C, -);
   for(int i = len1 - ; i < len2; ++ i)
     if(C[i].r < 0.5) {
       ans[++ cnt] = i - len1 + ;
     }
   printf("%d\n", cnt);
   for(int i = ; i <= cnt; ++ i) {
     printf("%d ", ans[i]);
   }

 #ifndef stone

   fclose(stdin); fclose(stdout);

 #endif

   return ;
 }

4259

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 =  + ;
 typedef long long ll;
 const int mod = 1e9 + ;
 #define Pi acos(-1)

 struct Cp {
   double r, i;
   Cp(double _r = , double _i = ) :
     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 << ];
 int n, m, l, r[N], lens;
 Cp a[N], b[N];
 ll Mp[N << ], t1[N], t2[N], p[N];

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

 ll Manacher(char *s, int len) {
   int l = ; ll res = ;
   Ma[l ++] = '$';
   Ma[l ++] = '#';
   for(int i = ; i < len; ++ i) {
     Ma[l ++] = s[i];
     Ma[l ++] = '#';
   }
   Ma[l] = '!';
   int Mx = , id = ;
   for(int i = ; i < l; ++ i) {
     if(Mx > i) {
       Mp[i] = min(Mp[ * id - i], (ll)Mx - i);
     }
     else {
       Mp[i] = ;
     }
     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] >> ); res %= mod;
   }
   return res;
 }

 int main() {
   ll ans = , tmp;
   scanf("%s", ss);
   n = strlen(ss); lens = strlen(ss);
   for(int i = ; i < n; ++ i) {
     if(ss[i] == 'a')
       a[i].r = , b[i].r = ;
     else a[i].r = , b[i].r = ;
   }
   m = (n << ) - ;
   for(n = ; n <= m; n <<= ) l ++;
   for(int i = ; i < n; ++ i) r[i] = (r[i >> ] >> ) |((i & ) << (l - ));
   DFT(a, ); DFT(b, );
   for(int i = ; i < n; ++ i) {
     a[i] = a[i] * a[i];
     b[i] = b[i] * b[i];
   }
   DFT(a, -); DFT(b, -);
   for(int i = ; i <= n; ++ i) {
     t1[i] = (ll) (a[i].r + 0.5);
     t2[i] = (ll) (b[i].r + 0.5);
   }
   for(int i = ; i < n; ++ i) {
     t1[i] += t2[i];
   }
   tmp = Manacher(ss, strlen(ss));
   p[] = ;
   for(int i = ; i <= lens; ++ i)
     p[i] = p[i - ] *  % mod;
   lens = lens + lens - ;
   for(int i = ; i < lens; ++ i) {
     if(i & ) {
       ans = (ans + p[t1[i] >> ] - ) % mod;
     }
     else {
       ans = (ans + (p[(t1[i] - ) >> ] << ) - ) % mod;
     }
   }
   ans = (ans - tmp) % mod;
   while(ans < ) {
     ans += mod;
     ans %= mod;
   }
   printf("%lld\n", ans);
   return ;
 }

万径人踪灭

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 =  + ;
 const int mod = ;
 typedef long long ll;

 ll n;

 inline ll read() {
   ll x = ;
   char c = getchar();
   while(!isdigit(c)) c = getchar();
   while(isdigit(c)) {
     x = x *  + c - '';
     x %= mod;
     c = getchar();
   }
   return x;
 }

 int main() {
   n = read();
   n = n * (n + ) * (n + );
   n /= ;
   n %= mod;
   printf("%lld\n", n);
   return ;
 }

3028