BZOJ3434 [Wc2014]时空穿梭

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 }
View Code(递推求组合数)

 

  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 }
View Code(奇怪的组合数求法)

(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。。。)

posted on 2015-02-16 22:58  Xs酱~  阅读(345)  评论(0)  编辑  收藏  举报