BZOJ 4806 - 4809 象棋四题

 1 #include<cstdio>
 2 #define N 105
 3 #define P 999983
 4 int n,m;long long f[N][N][N],ans;
 5 inline long long C(int a,int b){
 6     return b==1?a:(1LL*a*(a-1)/2)%P;
 7 }
 8 main(){
 9     scanf("%d%d",&n,&m),f[0][m][0]=1;
10     for(int i=0;i<n;++i)
11         for(int j=0;j<=m;++j)
12             for(int k=0;k<=m;++k)
13                 if(f[i][j][k]){
14                     (f[i+1][j][k]+=f[i][j][k])%=P;
15                     if(j>0)(f[i+1][j-1][k+1]+=f[i][j][k]*C(j,1))%=P;
16                     if(j>1)(f[i+1][j-2][k+2]+=f[i][j][k]*C(j,2))%=P;
17                     if(j>0)(f[i+1][j-1][k]+=f[i][j][k]*C(j,1)%P*C(k,1))%=P;
18                     if(k>0)(f[i+1][j][k-1]+=f[i][j][k]*C(k,1))%=P;
19                     if(k>1)(f[i+1][j][k-2]+=f[i][j][k]*C(k,2))%=P;
20                 }
21     for(int i=0;i<=m;++i)
22         for(int j=0;j<=m;++j)
23             ans=(ans+f[n][i][j])%P;
24     printf("%lld\n",ans);
25 }

 1 #include<cstdio>
 2 #define N 1000005
 3 int n,m,t,pri[N],que[N],cnt[N],ans[64],tot=1,mor;
 4 void calc(int lim,int flg){
 5     for(int i=2;i<=lim;++i)
 6         for(int num=i;num!=1;)
 7             cnt[pri[num]]+=flg,num/=pri[num];
 8 }
 9 main(){
10     scanf("%d%d",&n,&m);
11     if(n<m)n^=m^=n^=m;
12     for(int i=2;i<=n;++i){
13         if(!pri[i])que[++t]=pri[i]=i;
14         for(int j=1;j<=t&&i*que[j]<=n;++j){
15             pri[que[j]*i]=que[j];
16             if(!(i%que[j]))break;
17         }
18     }
19     ans[1]=1;
20     calc(n,+1);
21     calc(m,-1);
22     calc(n-m,-1);
23     for(int i=1;i<=t;++i)
24         for(int j=1;j<=cnt[que[i]];++j){tot+=7;
25             for(int k=1;k<=tot;++k)ans[k]*=que[i];
26             for(int k=1;k<=tot;++k)ans[k+1]+=ans[k]/10,ans[k]%=10;
27             while(tot&&!ans[tot])--tot;if(tot>50)tot=50;
28         }
29     while(tot&&!ans[tot])--tot;
30     for(int i=tot;i;--i)putchar(‘0‘+ans[i]);
31 }

  1 #include <cstdio>
  2  
  3 template <class T>
  4 inline T min(const T &a, const T &b)
  5 {
  6     return a < b ? a : b;
  7 }
  8  
  9 inline int nextInt(void)
 10 {
 11     int r = 0;
 12     int f = 0;
 13     int c = getchar();
 14      
 15     for (; c < 48; c = getchar())
 16         if (c == ‘-‘)f = 1;
 17      
 18     for (; c > 47; c = getchar())
 19         r = r * 10 + c - 48;
 20      
 21     return f ? -r : r;
 22 }
 23  
 24 const int mxn = 40005;
 25 const int mxm = 2000005;
 26 const int inf = 1000000007;
 27  
 28 int s;
 29 int t;
 30 int tt;
 31 int hd[mxn];
 32 int nt[mxm];
 33 int to[mxm];
 34 int fl[mxm];
 35  
 36 inline void add(int x, int y)
 37 {
 38     nt[tt] = hd[x], to[tt] = y, fl[tt] = 1, hd[x] = tt++;
 39     nt[tt] = hd[y], to[tt] = x, fl[tt] = 0, hd[y] = tt++;
 40 }
 41  
 42 int dep[mxn];
 43 int que[mxn];
 44 int cur[mxn];
 45  
 46 inline bool bfs(void)
 47 {
 48     for (int i = s; i <= t; ++i)
 49         dep[i] = 0;
 50      
 51     int l = 0, r = 1;
 52     dep[s] = 1;
 53     que[0] = s;
 54      
 55     while (l != r)
 56     {
 57         int u = que[l++], v;
 58          
 59         for (int i = hd[u]; ~i; i = nt[i])
 60             if (v = to[i], !dep[v] && fl[i])
 61                 dep[que[r++] = v] = dep[u] + 1;
 62     }
 63      
 64     return dep[t];
 65 }
 66  
 67 int dfs(int u, int f)
 68 {
 69     if (!f || u == t)
 70         return f;
 71      
 72     int used = 0, flow, v;
 73      
 74     for (int &i = cur[u]; ~i; i = nt[i])
 75         if (v = to[i], fl[i] && dep[v] == dep[u] + 1)
 76         {
 77             flow = dfs(v, min(f - used, fl[i]));
 78              
 79             used += flow;
 80             fl[i] -= flow;
 81             fl[i^1] += flow;
 82              
 83             if (used == f)
 84                 return used;
 85         }
 86      
 87     if (!used)
 88         dep[u] = 0;
 89      
 90     return used;
 91 }
 92  
 93 inline int flow(void)
 94 {
 95     int ret = 0, tmp;
 96      
 97     while (bfs())
 98     {
 99         for (int i = s; i <= t; ++i)
100             cur[i] = hd[i];
101          
102         while (tmp = dfs(s, inf))
103             ret += tmp;
104     }
105      
106     return ret;
107 }
108  
109 int n, m, id[205][205];
110  
111 const int mov[8][2] =
112 {
113     {+2, +1},
114     {+2, -1},
115     {-2, +1},
116     {-2, -1},
117     {+1, +2},
118     {+1, -2},
119     {-1, +2},
120     {-1, -2}
121 };
122  
123 signed main(void)
124 {
125     n = nextInt();
126     m = nextInt();
127      
128     for (int i = 1; i <= n; ++i)
129         for (int j = 1; j <= m; ++j)
130             if (!nextInt())id[i][j] = ++t;
131      
132     ++t;
133      
134     for (int i = s; i <= t; ++i)
135         hd[i] = -1;
136      
137     for (int i = 1; i <= n; ++i)
138         for (int j = 1; j <= m; ++j)
139             if (id[i][j])
140             {
141                 if ((i ^ j) & 1)
142                     add(s, id[i][j]);
143                 else
144                     add(id[i][j], t);
145             }
146      
147     for (int i = 1; i <= n; ++i)
148         for (int j = 1; j <= m; ++j)
149             if (((i ^ j) & 1) && id[i][j])
150                 for (int k = 0; k < 8; ++k)
151                 {
152                     int x = i + mov[k][0];
153                     int y = j + mov[k][1];
154                      
155                     if (x < 1 || x > n)continue;
156                     if (y < 1 || y > m)continue;
157                      
158                     if (!id[x][y])continue;
159                      
160                     add(id[i][j], id[x][y]);
161                 }
162      
163     printf("%d\n", t - 1 - flow());
164 }

 1 #include <cstdio>
 2 
 3 inline int nextInt(void)
 4 {
 5     int r = 0;
 6     int c = getchar();
 7     
 8     for (; c < 48; c = getchar());
 9     for (; c > 47; c = getchar())
10         r = r * 10 + c - 48;
11     
12     return r;
13 }
14 
15 const int mxn = 20;
16 
17 typedef unsigned long long lnt;
18 
19 int n, ans, map[mxn][mxn];
20 
21 void search(int x, lnt a, lnt b, lnt c)
22 {
23     if (x == n)++ans;
24     else
25     {
26         lnt d = a | b | c;
27         
28         for (register int y = 0; y < n; ++y)
29             if (!map[x][y] && !((d >> y) & 1))
30                 search(x + 1,
31                     (a | (1ULL << y)),
32                     (b | (1ULL << y)) >> 1,
33                     (c | (1ULL << y)) << 1);
34     }
35 }
36 
37 signed main(void)
38 {
39     n = nextInt();
40     
41     for (int i = 0; i < n; ++i)
42         for (int j = 0; j < n; ++j)
43             map[i][j] = nextInt();
44     
45     search(0, 0, 0, 0);
46     
47     printf("%d\n", ans);
48 }