摔电脑摔电脑！JZP业界毒瘤！
400题纪念~哇终于上400了的说！！！好不容易欸！
题解什么的还是Orz iwtwiioi
我求组合数的方法明明是O(n)的，为什么这么慢！！！令人报警！
喂，话说这题的重点不在求组合数上面吧。。。
1 /************************************************************** 2 Problem: 3434 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:6888 ms 7 Memory:140824 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <cstring> 12 #include <algorithm> 13 14 using namespace std; 15 const int N = 100005; 16 const int mod = 10007; 17 const int Tt = 1005; 18 19 int g[21][N], f[21][15][N]; 20 int a[15], p[N], cnt, u[N], mx, mn; 21 int n[Tt], c[Tt], m[Tt][15]; 22 int C[N][20]; 23 int ans; 24 bool vis[N]; 25 26 inline int read() { 27 int x = 0, sgn = 1; 28 char ch = getchar(); 29 while (ch < '0' || '9' < ch) { 30 if (ch == '-') sgn = -1; 31 ch = getchar(); 32 } 33 while ('0' <= ch && ch <= '9') { 34 x = x * 10 + ch - '0'; 35 ch = getchar(); 36 } 37 return sgn * x; 38 } 39 40 int pow(int x, int y) { 41 int res = 1; 42 while (y) { 43 if (y & 1) (res *= x) %= mod; 44 (x *= x) %= mod; 45 y >>= 1; 46 } 47 return res; 48 } 49 50 void get_f(int N) { 51 int i, j, k, l; 52 for (u[1] = 1, i = 2; i <= N; ++i) { 53 if (!vis[i]) 54 p[++cnt] = i, u[i] = -1; 55 for (j = 1; j <= cnt; ++j) { 56 if ((k = i * p[j]) > N) break; 57 vis[k] = 1; 58 if (i % p[j] == 0) { 59 u[k] = 0; 60 break; 61 } 62 u[k] = -u[i]; 63 } 64 } 65 for (C[0][0] = i = 1; i <= N; ++i) 66 for (C[i][0] = j = 1; j < 19; ++j) 67 C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % mod; 68 69 for (l = 2; l <= 20; ++l) 70 for (i = 1; i <= N; ++i) 71 for (j = i, k = 1; j <= N; j += i, ++k) 72 (g[l][j] += C[i - 1][l - 2] * u[k]) %= mod; 73 for (i = 1; i <= N; ++i) 74 for (l = 2; l <= 20; ++l) { 75 k = 1; 76 for (j = 1; j <= 12; ++j) { 77 f[l][j][i] = (f[l][j][i - 1] + k * g[l][i]) % mod; 78 (k *= i) %= mod; 79 } 80 } 81 } 82 83 void calc(int i, int x) { 84 memset(a, 0, sizeof(a)); 85 a[1] = 1; 86 int d1, d2, t, j, k; 87 for (k = 1; k <= n[i]; ++k) { 88 t = m[i][k] / x; 89 d1 = (1ll * t * m[i][k]) % mod; 90 d2 = (1ll * t * (t + 1) >> 1) % mod; 91 for (j = n[i] + 1; j > 1; --j) 92 a[j] =(a[j] * d1 - a[j - 1] * d2) % mod; 93 (a[1] *= d1) %= mod; 94 } 95 } 96 97 int main() { 98 int i, j, k, l, T; 99 T = read(); 100 for (i = 1; i <= T; ++i) { 101 n[i] = read(), c[i] = read(); 102 for (j = 1; j <= n[i]; ++j) 103 m[i][j] = read(), mx = max(mx, m[i][j]); 104 } 105 get_f(mx); 106 107 for (i = 1; i <= T; ++i) { 108 ans = 0; 109 for (mn = m[i][1], j = 2; j <= n[i]; ++j) 110 mn = min(mn, m[i][j]); 111 for (j = 1; j <= mn; j = k + 1) { 112 k = mn + 1; 113 for (l = 1; l <= n[i]; ++l) 114 k = min(k, m[i][l] / (m[i][l] / j)); 115 calc(i, j); 116 for (l = 1; l <= n[i] + 1; ++l) 117 (ans += a[l] * (f[c[i]][l][k] - f[c[i]][l][j - 1])) %= mod; 118 } 119 ans %= mod; 120 printf("%d\n", (ans += mod) %= mod); 121 } 122 return 0; 123 }
1 /************************************************************** 2 Problem: 3434 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:9596 ms 7 Memory:134184 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <cstring> 12 #include <algorithm> 13 14 using namespace std; 15 const int N = 100005; 16 const int mod = 10007; 17 const int Tt = 1005; 18 19 int g[21][N], f[21][15][N]; 20 int a[15], p[N], cnt, u[N], mx, mn; 21 int n[Tt], c[Tt], m[Tt][15]; 22 int fac[N], faci[N], sum[N]; 23 int ans; 24 bool vis[N]; 25 26 inline int read() { 27 int x = 0, sgn = 1; 28 char ch = getchar(); 29 while (ch < '0' || '9' < ch) { 30 if (ch == '-') sgn = -1; 31 ch = getchar(); 32 } 33 while ('0' <= ch && ch <= '9') { 34 x = x * 10 + ch - '0'; 35 ch = getchar(); 36 } 37 return sgn * x; 38 } 39 40 int pow(int x, int y) { 41 int res = 1; 42 while (y) { 43 if (y & 1) (res *= x) %= mod; 44 (x *= x) %= mod; 45 y >>= 1; 46 } 47 return res; 48 } 49 50 inline int C(int x, int y) { 51 if (y == 0) return 1; 52 if (x < y) return 0; 53 if (sum[x] != sum[y] + sum[x - y]) return 0; 54 return fac[x] * faci[y] % mod * faci[x - y] % mod; 55 } 56 57 void get_f(int N) { 58 int i, j, k, l; 59 for (u[1] = 1, i = 2; i <= N; ++i) { 60 if (!vis[i]) 61 p[++cnt] = i, u[i] = -1; 62 for (j = 1; j <= cnt; ++j) { 63 if ((k = i * p[j]) > N) break; 64 vis[k] = 1; 65 if (i % p[j] == 0) { 66 u[k] = 0; 67 break; 68 } 69 u[k] = -u[i]; 70 } 71 } 72 fac[0] = faci[0] = 1; 73 for (fac[1] = 1, i = 2; i <= N; ++i) 74 if (i % mod == 0) { 75 sum[i] = sum[i - 1] + 1; 76 fac[i] = fac[i - 1] * (i / mod) % mod; 77 } else { 78 sum[i] = sum[i - 1]; 79 fac[i] = fac[i - 1] * i % mod; 80 } 81 for (faci[N] = pow(fac[N], mod - 2),i = N - 1; i; --i) 82 if ((i + 1) % mod == 0) 83 faci[i] = faci[i + 1] * ((i + 1) / mod) % mod; 84 else faci[i] = faci[i + 1] * (i + 1) % mod; 85 86 for (l = 2; l <= 20; ++l) 87 for (i = 1; i <= N; ++i) 88 for (j = i, k = 1; j <= N; j += i, ++k) 89 (g[l][j] += C(i - 1, l - 2) * u[k]) %= mod; 90 for (i = 1; i <= N; ++i) 91 for (l = 2; l <= 20; ++l) { 92 k = 1; 93 for (j = 1; j <= 12; ++j) { 94 f[l][j][i] = (f[l][j][i - 1] + k * g[l][i]) % mod; 95 (k *= i) %= mod; 96 } 97 } 98 } 99 100 void calc(int i, int x) { 101 memset(a, 0, sizeof(a)); 102 a[1] = 1; 103 int d1, d2, t, j, k; 104 for (k = 1; k <= n[i]; ++k) { 105 t = m[i][k] / x; 106 d1 = (1ll * t * m[i][k]) % mod; 107 d2 = (1ll * t * (t + 1) >> 1) % mod; 108 for (j = n[i] + 1; j > 1; --j) 109 a[j] =(a[j] * d1 - a[j - 1] * d2) % mod; 110 (a[1] *= d1) %= mod; 111 } 112 } 113 114 int main() { 115 int i, j, k, l, T; 116 T = read(); 117 for (i = 1; i <= T; ++i) { 118 n[i] = read(), c[i] = read(); 119 for (j = 1; j <= n[i]; ++j) 120 m[i][j] = read(), mx = max(mx, m[i][j]); 121 } 122 get_f(mx); 123 124 for (i = 1; i <= T; ++i) { 125 ans = 0; 126 for (mn = m[i][1], j = 2; j <= n[i]; ++j) 127 mn = min(mn, m[i][j]); 128 for (j = 1; j <= mn; j = k + 1) { 129 k = mn + 1; 130 for (l = 1; l <= n[i]; ++l) 131 k = min(k, m[i][l] / (m[i][l] / j)); 132 calc(i, j); 133 for (l = 1; l <= n[i] + 1; ++l) 134 (ans += a[l] * (f[c[i]][l][k] - f[c[i]][l][j - 1])) %= mod; 135 } 136 ans %= mod; 137 printf("%d\n", (ans += mod) %= mod); 138 } 139 return 0; 140 }
(p.s. 关于那种空间较小的求组合数的方法为什么慢了3秒我也是无言以对。。。)
