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
1 #include <cstdlib> 2 #include <iostream> 3 #include <algorithm> 4 #include <cstring> 5 #include <cstdio> 6 #include <cmath> 7 #define Pi acos(-1) 8 9 using namespace std; 10 const int N = 600000 + 5; 11 const int rad = 10; 12 13 struct Complex { 14 double r, i; 15 Complex(double x = 0, double y = 0) { r = x; i = y; } 16 Complex operator + (Complex x) { return Complex(r + x.r, i + x.i); } 17 Complex operator - (Complex x) { return Complex(r - x.r, i - x.i); } 18 Complex operator * (Complex x) { return Complex(r * x. r - i * x.i, r * x.i + i * x.r); } 19 }a[N], b[N];; 20 21 int n, m, l, r[N], c[N]; 22 char ss[N]; 23 24 void dft(Complex *s, int f) { 25 for(int i = 0; i < n; ++ i) 26 if(i < r[i]) swap(s[i], s[r[i]]); 27 for(int i = 1; i < n; i <<= 1) { 28 Complex wn(cos(Pi / i), f * sin(Pi / i)); 29 for(int j = 0; j < n; j += (i << 1)) { 30 Complex w(1, 0); 31 for(int k = 0; k < i; ++ k) { 32 Complex x = s[j + k], y = w * s[j + k + i]; 33 s[j + k] = x + y; 34 s[j + k + i] = x - y; 35 w = w * wn; 36 } 37 } 38 } 39 if(f == -1) 40 for(int i = 0; i < n; ++ i) 41 s[i].r /= n; 42 } 43 44 int main() { 45 #ifndef stone 46 freopen("bettermul.in", "r", stdin); 47 freopen("bettermul.out", "w", stdout); 48 #endif 49 50 int len1, len2; 51 scanf("%s", ss); len1 = strlen(ss); 52 for(int i = 0; i < len1; ++ i) a[i] = ss[len1 - i - 1] - '0'; 53 scanf("%s", ss); len2 = strlen(ss); 54 for(int i = 0; i < len2; ++ i) b[i] = ss[len2 - i - 1] - '0'; 55 n = max(len1, len2); m = (n << 1) - 2; 56 for(n = 1; n <= m; n <<= 1) l ++; 57 for(int i = 0; i < n; ++ i) 58 r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1)); 59 dft(a, 1); dft(b, 1); 60 for(int i = 0; i < n; ++ i) a[i] = a[i] * b[i]; 61 dft(a, -1); 62 for(int i = 0; i <= m; ++ i) c[i] = (int) (a[i].r + 0.5); 63 for(int i = 0; i <= m; ++ i) { 64 if(c[i] >= rad) { 65 c[i + 1] += c[i] / rad; 66 c[i] %= rad; 67 if(i == m) m ++; 68 } 69 } 70 int pos = m; 71 while(c[pos] == 0) -- pos; 72 for(int i = pos; i >= 0; -- i) printf("%d", c[i]); 73 74 #ifndef stone 75 fclose(stdin); fclose(stdout); 76 #endif 77 return 0; 78 }
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。
1 #include <iostream> 2 #include <algorithm> 3 #include <cstdlib> 4 #include <cstring> 5 #include <cstdio> 6 #include <cmath> 7 8 using namespace std; 9 typedef long long ll; 10 const int N = 600000 + 5; 11 #define Pi acos(-1) 12 13 struct Cp { 14 double r, i; 15 Cp(double _r = 0, double _i = 0): 16 r(_r), i(_i) {} 17 Cp operator + (Cp x) { 18 return Cp(r + x.r, i + x.i); 19 } 20 Cp operator - (Cp x) { 21 return Cp(r - x.r, i - x.i); 22 } 23 Cp operator * (Cp x) { 24 return Cp(r * x.r - i * x.i, r * x.i + i * x.r); 25 } 26 }; 27 28 char str1[N], str2[N]; 29 int len1, len2, n, m, l, cnt; 30 int t[N], ans[N], r[N]; 31 Cp a[N], b[N], A[N], B[N], C[N], tp; 32 33 34 void DFT(Cp *s, int f) { 35 for(int i = 0; i < n; ++ i) 36 if(i < r[i]) swap(s[i], s[r[i]]); 37 for(int i = 1; i < n; i <<= 1) { 38 Cp wn(cos(Pi / i), f * sin(Pi / i)); 39 for(int j = 0; j < n; j += (i << 1)) { 40 Cp w(1, 0); 41 for(int k = 0; k < i; ++ k) { 42 Cp x = s[j + k], y = w * s[j + k + i]; 43 s[j + k] = x + y; 44 s[j + k + i] = x - y; 45 w = w * wn; 46 } 47 } 48 } 49 if(f == -1) 50 for(int i = 0; i < n; ++ i) 51 s[i].r /= n; 52 } 53 54 #define stone 55 56 int main() { 57 #ifndef stone 58 59 freopen("bitch.in", "r", stdin); 60 freopen("bitch.out", "w", stdout); 61 62 #endif 63 64 scanf("%d%d", &len1, &len2); 65 scanf("%s%s", str1, str2); 66 reverse(str1, str1 + len1); 67 for(int i = 0; i < len1; ++ i) { 68 if(str1[i] == '*') a[i].r = 0; 69 else a[i].r = (str1[i] - 'a' + 1); 70 } 71 for(int i = 0; i < len2; ++ i) { 72 if(str2[i] == '*') b[i].r = 0; 73 else b[i].r = (str2[i] - 'a' + 1); 74 } 75 m = len1 + len2; 76 for(n = 1; n <= m; n <<= 1) ++ l; 77 for(int i = 0; i < n; ++ i) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1)); 78 79 for(int i = 0; i < n; ++ i) { 80 A[i].r = a[i].r * a[i].r * a[i].r; 81 B[i].r = b[i].r; 82 } 83 DFT(A, 1); DFT(B, 1); 84 for(int i = 0; i < n; ++ i) { C[i] = C[i] + A[i] * B[i]; } 85 memset(A, 0, sizeof A); memset(B, 0, sizeof B); 86 for(int i = 0; i < n; ++ i) { 87 A[i].r = a[i].r * a[i].r; 88 B[i].r = b[i].r * b[i].r; 89 } 90 DFT(A, 1); DFT(B, 1); tp = (Cp) {2, 0}; 91 for(int i = 0; i < n; ++ i) { C[i] = C[i] - A[i] * B[i] * tp; } 92 memset(A, 0, sizeof A); memset(B, 0, sizeof B); 93 for(int i = 0; i < n; ++ i) { 94 A[i].r = a[i].r; 95 B[i].r = b[i].r * b[i].r * b[i].r; 96 } 97 DFT(A, 1); DFT(B, 1); 98 for(int i = 0; i < n; ++ i) { C[i] = C[i] + A[i] * B[i]; } 99 DFT(C, -1); 100 for(int i = len1 - 1; i < len2; ++ i) 101 if(C[i].r < 0.5) { 102 ans[++ cnt] = i - len1 + 2; 103 } 104 printf("%d\n", cnt); 105 for(int i = 1; i <= cnt; ++ i) { 106 printf("%d ", ans[i]); 107 } 108 109 #ifndef stone 110 111 fclose(stdin); fclose(stdout); 112 113 #endif 114 115 return 0; 116 }
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。
1 #include <cstdlib> 2 #include <algorithm> 3 #include <cstdio> 4 #include <cstring> 5 #include <iostream> 6 #include <cmath> 7 8 using namespace std; 9 const int N = 400000 + 5; 10 typedef long long ll; 11 const int mod = 1e9 + 7; 12 #define Pi acos(-1) 13 14 struct Cp { 15 double r, i; 16 Cp(double _r = 0, double _i = 0) : 17 r(_r), i(_i){} 18 Cp operator + (Cp x) { 19 return Cp(r + x.r, i + x.i); 20 } 21 Cp operator - (Cp x) { 22 return Cp(r - x.r, i - x.i); 23 } 24 Cp operator * (Cp x) { 25 return Cp(r * x.r - i * x.i, r * x.i + i * x.r); 26 } 27 }; 28 29 char ss[N], Ma[N << 1]; 30 int n, m, l, r[N], lens; 31 Cp a[N], b[N]; 32 ll Mp[N << 1], t1[N], t2[N], p[N]; 33 34 void DFT(Cp *s, int f) { 35 for(int i = 0; i < n; ++ i) 36 if(i < r[i]) swap(s[i], s[r[i]]); 37 for(int i = 1; i < n; i <<= 1) { 38 Cp wn(cos(Pi / i), f * sin(Pi / i)); 39 for(int j = 0; j < n; j += (i << 1)) { 40 Cp w(1, 0); 41 for(int k = 0; k < i; ++ k) { 42 Cp x = s[j + k], y = w * s[j + k + i]; 43 s[j + k] = x + y; 44 s[j + k + i] = x - y; 45 w = w * wn; 46 } 47 } 48 } 49 if(f == -1) 50 for(int i = 0; i < n; ++ i) 51 s[i].r /= n; 52 } 53 54 ll Manacher(char *s, int len) { 55 int l = 0; ll res = 0; 56 Ma[l ++] = '$'; 57 Ma[l ++] = '#'; 58 for(int i = 0; i < len; ++ i) { 59 Ma[l ++] = s[i]; 60 Ma[l ++] = '#'; 61 } 62 Ma[l] = '!'; 63 int Mx = 0, id = 0; 64 for(int i = 0; i < l; ++ i) { 65 if(Mx > i) { 66 Mp[i] = min(Mp[2 * id - i], (ll)Mx - i); 67 } 68 else { 69 Mp[i] = 1; 70 } 71 while(Ma[i + Mp[i]] == Ma[i - Mp[i]]) Mp[i] ++; 72 if(i + Mp[i] > Mx) { 73 Mx = i + Mp[i]; 74 id = i; 75 } 76 res += (Mp[i] >> 1); res %= mod; 77 } 78 return res; 79 } 80 81 int main() { 82 ll ans = 0, tmp; 83 scanf("%s", ss); 84 n = strlen(ss); lens = strlen(ss); 85 for(int i = 0; i < n; ++ i) { 86 if(ss[i] == 'a') 87 a[i].r = 1, b[i].r = 0; 88 else a[i].r = 0, b[i].r = 1; 89 } 90 m = (n << 1) - 2; 91 for(n = 1; n <= m; n <<= 1) l ++; 92 for(int i = 0; i < n; ++ i) r[i] = (r[i >> 1] >> 1) |((i & 1) << (l - 1)); 93 DFT(a, 1); DFT(b, 1); 94 for(int i = 0; i < n; ++ i) { 95 a[i] = a[i] * a[i]; 96 b[i] = b[i] * b[i]; 97 } 98 DFT(a, -1); DFT(b, -1); 99 for(int i = 0; i <= n; ++ i) { 100 t1[i] = (ll) (a[i].r + 0.5); 101 t2[i] = (ll) (b[i].r + 0.5); 102 } 103 for(int i = 0; i < n; ++ i) { 104 t1[i] += t2[i]; 105 } 106 tmp = Manacher(ss, strlen(ss)); 107 p[0] = 1; 108 for(int i = 1; i <= lens; ++ i) 109 p[i] = p[i - 1] * 2 % mod; 110 lens = lens + lens - 1; 111 for(int i = 0; i < lens; ++ i) { 112 if(i & 1) { 113 ans = (ans + p[t1[i] >> 1] - 1) % mod; 114 } 115 else { 116 ans = (ans + (p[(t1[i] - 1) >> 1] << 1) - 1) % mod; 117 } 118 } 119 ans = (ans - tmp) % mod; 120 while(ans < 0) { 121 ans += mod; 122 ans %= mod; 123 } 124 printf("%lld\n", ans); 125 return 0; 126 }
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。。。。。
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <cstdlib> 5 #include <algorithm> 6 #include <cctype> 7 8 using namespace std; 9 const int N = 500 + 5; 10 const int mod = 10007; 11 typedef long long ll; 12 13 ll n; 14 15 inline ll read() { 16 ll x = 0; 17 char c = getchar(); 18 while(!isdigit(c)) c = getchar(); 19 while(isdigit(c)) { 20 x = x * 10 + c - '0'; 21 x %= mod; 22 c = getchar(); 23 } 24 return x; 25 } 26 27 int main() { 28 n = read(); 29 n = n * (n + 1) * (n + 2); 30 n /= 6; 31 n %= mod; 32 printf("%lld\n", n); 33 return 0; 34 }
