NOIP模拟73
T1：
20分：状压DP或Dfs O(2^n)暴力
60分：发现答案为1e5级别，做0/1背包即可
100分：分析过程，由于要求最小的非幸运数，那么由于数的组成是连续的
那么模拟过程发现，能够延伸所能组成的数的集合的单位数为：1,2,4,11,22,44
进一步拓展发现第i个数为Pre[i - 1] + 1，因此若a[i] > Pre[i - 1] + 1，那么答案则为
Pre[i - 1] + 1，时间复杂度O(nlogn)（sort排序）
代码如下：
1 #include <bits/stdc++.h> 2 using namespace std; 3 #define I long long 4 #define C char 5 #define B bool 6 #define V void 7 #define D double 8 #define LL long long 9 #define UI unsigned int 10 #define UL unsigned long long 11 #define P pair<I,I> 12 #define MP make_pair 13 #define a first 14 #define b second 15 #define lowbit(x) (x & -x) 16 #define debug cout << "It's Ok Here !" << endl; 17 #define FP(x) freopen (#x,"r",stdin) 18 #define FC(x) freopen (#x,"w",stdout) 19 const I N = 1e5 + 3; 20 I n,a[N],Pre[N]; 21 inline I read () { 22 I x(0),y(1); C z(getchar()); 23 while (!isdigit(z)) { if (z == '-') y = -1; z = getchar(); } 24 while ( isdigit(z)) x = x * 10 + (z ^ 48), z = getchar(); 25 return x * y; 26 } 27 inline V Max (I &a,I b) { a = a > b ? a : b; } 28 inline V Min (I &a,I b) { a = a < b ? a : b; } 29 inline I max (I a,I b) { return a > b ? a : b; } 30 inline I min (I a,I b) { return a < b ? a : b; } 31 inline V swap (I &a,I &b) { a ^= b, b ^= a, a ^= b; } 32 inline I abs (I a) { return a >= 0 ? a : -a; } 33 inline P operator + (const P &a,const P &b) { 34 return MP (a.a + b.a,a.b + b.b); 35 } 36 inline P operator - (const P &a,const P &b) { 37 return MP (a.a - b.a,a.b - b.b); 38 } 39 40 signed main () { 41 FP (math.in), FC (math.out); 42 n = read (); 43 for (I i(1);i <= n; ++ i) 44 a[i] = read (); 45 sort (a + 1,a + n + 1); 46 for (I i(1);i <= n; ++ i) 47 Pre[i] = Pre[i - 1] + a[i]; 48 for (I i(1);i <= n; ++ i) 49 if (a[i] > Pre[i - 1] + 1) 50 printf ("%lld\n",Pre[i - 1] + 1), exit (0); 51 printf ("%lld\n",Pre[n] + 1), exit (0); 52 }
T2：
矩阵乘法，重新定义矩阵乘法为C[i][j] = $\prod$A[i][k]^B[k][j]，于是直接调配矩阵系数即可
需要注意的是，由于将幂乘转化为指数相加，那么在指数矩阵取模时需要根据费马定理对(mod - 1)取模
代码如下：
1 %:pragma GCC optimize (2) 2 #include <bits/stdc++.h> 3 using namespace std; 4 #define I long long 5 #define C char 6 #define B bool 7 #define V void 8 #define D double 9 #define LL long long 10 #define UI unsigned int 11 #define UL unsigned long long 12 #define P pair<I,I> 13 #define MP make_pair 14 #define a first 15 #define b second 16 #define lowbit(x) (x & -x) 17 #define debug cout << "It's Ok Here !" << endl; 18 #define FP(x) freopen (#x,"r",stdin) 19 #define FC(x) freopen (#x,"w",stdout) 20 const I N = 200, mod = 998244353; 21 I n,t,F[N],G[N][N]; 22 inline I read () { 23 I x(0),y(1); C z(getchar()); 24 while (!isdigit(z)) { if (z == '-') y = -1; z = getchar(); } 25 while ( isdigit(z)) x = x * 10 + (z ^ 48), z = getchar(); 26 return x * y; 27 } 28 inline V Max (I &a,I b) { a = a > b ? a : b; } 29 inline V Min (I &a,I b) { a = a < b ? a : b; } 30 inline I max (I a,I b) { return a > b ? a : b; } 31 inline I min (I a,I b) { return a < b ? a : b; } 32 inline V swap (I &a,I &b) { a ^= b, b ^= a, a ^= b; } 33 inline I abs (I a) { return a >= 0 ? a : -a; } 34 inline P operator + (const P &a,const P &b) { 35 return MP (a.a + b.a,a.b + b.b); 36 } 37 inline P operator - (const P &a,const P &b) { 38 return MP (a.a - b.a,a.b - b.b); 39 } 40 inline I fp (I a,I b) { 41 if (a == 0 || a == 1) return a; 42 I ans (1); 43 for (; b ;b >>= 1, a = a * a % mod) 44 if (b & 1) ans = ans * a % mod; 45 return ans; 46 } 47 inline V Matrix_mul () { 48 I X[N]; 49 fill (X,X + t,1); 50 for (I k(0);k < t; ++ k) 51 for (I j(0);j < t; ++ j) 52 X[j] = X[j] * fp (F[k],G[k][j]) % mod; 53 memcpy (F,X,sizeof F); 54 } 55 inline V Matrix_self () { 56 I X[N][N]; 57 memset (X,0,sizeof X); 58 for (I i(0);i < t; ++ i) 59 for (I k(0);k < t; ++ k) if (G[i][k]) 60 for (I j(0);j < t; ++ j) if (G[k][j]) 61 (X[i][j] += G[i][k] * G[k][j] % (mod - 1)) %= (mod - 1); 62 memcpy (G,X,sizeof X); 63 } 64 signed main () { 65 FP (seq.in), FC (seq.out); 66 n = read (), t = read (); 67 for (I i(0);i < t; ++ i) 68 G[i][0] = read (); 69 for (I i(0);i < t; ++ i) 70 F[t - i - 1] = read (); 71 for (I i(0);i + 1 < t; ++ i) 72 G[i][i + 1] = 1; 73 if (n <= t) printf ("%lld\n",F[t - n]), exit (0); 74 I x (n - t); 75 for (; x ;x >>= 1, Matrix_self ()) 76 if (x & 1) Matrix_mul (); 77 printf ("%lld\n",F[0]); 78 }
T3：
全源最短路，考虑在最短路过程中记录每个白点连向最短路的连边，最终在每个白点所有连边集合中取最小值即可
代码如下：
1 #include <bits/stdc++.h> 2 using namespace std; 3 #define I long long 4 #define C char 5 #define B bool 6 #define V void 7 #define LL long long 8 #define UI unsigned int 9 #define UL unsigned long long 10 #define P pair<I,I> 11 #define MP make_pair 12 #define a first 13 #define b second 14 #define lowbit(x) (x & -x) 15 #define debug cout << "It's Ok Here !" << endl; 16 #define FP(x) freopen (#x,"r",stdin) 17 #define FC(x) freopen (#x,"w",stdout) 18 const I N = 1e5 + 3; 19 B typ[N]; 20 I n,m,d[N],pre[N],ans; 21 I tot,head[N],to[N << 2],nxt[N << 2],wgt[N << 2]; 22 vector <I> rec[N]; 23 inline I read () { 24 I x(0),y(1); C z(getchar()); 25 while (!isdigit(z)) { if (z == '-') y = -1; z = getchar(); } 26 while ( isdigit(z)) x = x * 10 + (z ^ 48), z = getchar(); 27 return x * y; 28 } 29 inline V Max (I &a,I b) { a = a > b ? a : b; } 30 inline V Min (I &a,I b) { a = a < b ? a : b; } 31 inline I max (I a,I b) { return a > b ? a : b; } 32 inline I min (I a,I b) { return a < b ? a : b; } 33 inline V swap (I &a,I &b) { a ^= b, b ^= a, a ^= b; } 34 inline I abs (I a) { return a >= 0 ? a : -a; } 35 inline P operator + (const P &a,const P &b) { 36 return MP (a.a + b.a,a.b + b.b); 37 } 38 inline P operator - (const P &a,const P &b) { 39 return MP (a.a - b.a,a.b - b.b); 40 } 41 inline V found (I z,I y,I x) { 42 to[++tot] = y, nxt[tot] = head[x], wgt[tot] = z, head[x] = tot; 43 to[++tot] = x, nxt[tot] = head[y], wgt[tot] = z, head[y] = tot; 44 } 45 inline V Dijkstra () { 46 priority_queue <P> q; B vis[N]; 47 memset (vis,0,sizeof vis), memset (d,0x3f,sizeof d); 48 for (I i(1);i <= n; ++ i) if (typ[i]) 49 q.push (MP (0,i)), d[i] = 0; 50 while (!q.empty ()) { 51 I x (q.top ().b); q.pop (); 52 if (vis[x]) continue; vis[x] = 1; 53 for (I i(head[x]),y(to[i]),z(wgt[i]); i ;i = nxt[i],y = to[i],z = wgt[i]) { 54 if (d[x] + z == d[y]) rec[y].push_back (z); 55 if (d[x] + z < d[y]) d[y] = d[x] + z, q.push (MP (-d[y],y)), rec[y].clear (), rec[y].push_back (z); 56 } 57 } 58 } 59 signed main () { 60 FP (minimum.in), FC (minimum.out); 61 n = read (), m = read (); 62 for (I i(1);i <= n; ++ i) 63 typ[i] = read (); 64 for (I i(1);i <= m; ++ i) 65 found (read (),read (),read ()); 66 Dijkstra (); 67 for (I i(1);i <= n; ++ i) if (!typ[i]) { 68 if (rec[i].empty ()) puts ("impossible"), exit (0); 69 I MI (INT_MAX); for (auto x : rec[i]) Min (MI,x); 70 ans += MI; 71 } 72 printf ("%lld\n",ans); 73 }
T4：
状压DP，是状态划分类型，考虑定义f[i]表示只考虑i集合中所有元素的拓扑序方案数
那么对于(1 << j & i == 0)f[i | 1 << j] += f[i] * (1 << cnt)，其中j为非i集合中的点，cnt为
i集合中连向j点的边数，其意义为考虑将j点作为新集合的拓扑序最后一个，那么i集合对新集合
的贡献即为f[i] * (1 << cnt)，即任意删边
代码如下：
1 #include <bits/stdc++.h> 2 using namespace std; 3 #define I long long 4 #define C char 5 #define B bool 6 #define V void 7 #define D double 8 #define LL long long 9 #define UI unsigned int 10 #define UL unsigned long long 11 #define P pair<I,I> 12 #define MP make_pair 13 #define a first 14 #define b second 15 #define lowbit(x) (x & -x) 16 #define debug cout << "It's Ok Here !" << endl; 17 #define FP(x) freopen (#x,"r",stdin) 18 #define FC(x) freopen (#x,"w",stdout) 19 const I N = 22, mod = 998244353; 20 I n,m,ans,f[1 << N],s[N]; 21 I tot,head[N],to[N*N],nxt[N*N]; 22 inline I read () { 23 I x(0),y(1); C z(getchar()); 24 while (!isdigit(z)) { if (z == '-') y = -1; z = getchar(); } 25 while ( isdigit(z)) x = x * 10 + (z ^ 48), z = getchar(); 26 return x * y; 27 } 28 inline V Max (I &a,I b) { a = a > b ? a : b; } 29 inline V Min (I &a,I b) { a = a < b ? a : b; } 30 inline I max (I a,I b) { return a > b ? a : b; } 31 inline I min (I a,I b) { return a < b ? a : b; } 32 inline V swap (I &a,I &b) { a ^= b, b ^= a, a ^= b; } 33 inline I abs (I a) { return a >= 0 ? a : -a; } 34 inline P operator + (const P &a,const P &b) { 35 return MP (a.a + b.a,a.b + b.b); 36 } 37 inline P operator - (const P &a,const P &b) { 38 return MP (a.a - b.a,a.b - b.b); 39 } 40 inline V found (I y,I x) { 41 to[++tot] = y, nxt[tot] = head[x], head[x] = y, s[y] |= 1 << x; 42 } 43 signed main () { 44 FP (topology.in), FC (topology.out); 45 n = read (), m = read (); 46 for (I i(1);i <= m; ++ i) 47 found (read () - 1, read () - 1); 48 f[0] = 1; 49 for (I i(0);i < 1 << n; ++ i) 50 for (I j(0);j < n; ++ j) if ((1 << j & i) == 0) { 51 I cnt (__builtin_popcount (s[j] & i)); 52 (f[i | 1 << j] += f[i] * (1 << cnt) % mod) %= mod; 53 } 54 printf ("%lld\n",f[(1 << n) - 1]); 55 }
附加题：奇妙的 Fibonacci
结论题+积性筛模板：
对于Fibonacci数列有：
通项公式：
证明：数学归纳法或者数列通项公式计算公式
恒等式：
证明：1~4通过fib定义式展开化简即可
5：通过fib展开观察得到：
f[n] = f[n - 1] + f[n - 2]
= 2 * f[n - 2] + f[n - 3]
= 3 * f[n - 3] + 2 * f[n - 4]
数论性质：
证明：首先证明fib相邻两项互质：
比较简单的数学归纳法：(a,b) = 1 -> (a,a + b) = 1
那么由上述恒等式5可得：证明式等价于：
(f[m] * f[n - m + 1] + f[m - 1] * f[n - m],f[m]) = f[(n,m)]
简单的数论化简得到(f[n - m],f[m]) = f[(n,m)]
可以发现其为辗转相除形式，由于最终一定存在：
(0,(a,b)) = (a,b)，也就是(0,f[(n,m)]) = f[(n,m)] Q.E.D
性质2：
有了上述证明，本性质比较显然，考虑若f[n] | f[m]
那么(f[n],f[m]) = f[n] = f[(n,m)]，那么可以的得到：
n = (n,m) -> n | m
于是本题直接积性筛预处理即可
（约数平方和积性函数证明：根据公式，对于每一项做平方即可）
火羽白日生的代码：
1 #include <bits/stdc++.h> 2 #define ll long long 3 #define ull unsigned long long 4 #define uint unsigned int 5 #define rint register int 6 #define int long long 7 using namespace std; 8 namespace IO{ 9 #define File(x,y) freopen(#x,"r",stdin),freopen(#y,"w",stdout) 10 #define debug puts("debug") 11 #define breakpoint exit(0) 12 #define out(x) cout<<#x<<" : "<<x<<"\n" 13 inline int read(){ 14 int w=0,f=1; char ch=getchar(); 15 while(ch<'0'||ch>'9'){if(ch=='-') f=-1; ch=getchar();} 16 while(ch>='0'&&ch<='9'){w=(w<<3)+(w<<1)+(ch^48); ch=getchar();} 17 return w*f; 18 } 19 inline void write(int x,char ch='\n'){ 20 static int sta[35]; int top=0; 21 if(x<0) putchar('-'),x=-x; 22 do{sta[++top]=x%10,x/=10;}while(x); 23 while(top) putchar(sta[top--]+48); putchar(ch); 24 } 25 } 26 using namespace IO; 27 namespace CL{ 28 #define fill(x,y) memset(x,y,sizeof(x)) 29 #define copy(x,y) memcpy(x,y,sizeof(y)) 30 31 const int maxn=1e7+5,mod=1e9+7; 32 33 int q,qi,A,B,C,tot,ans1,ans2; 34 int prime[maxn],d[maxn],sqr[maxn],low[maxn]; 35 bool is[maxn]; 36 inline void init(int n){ 37 fill(is,1); 38 d[1]=sqr[1]=1; 39 for(int i=2;i<=n;i++){ 40 if(is[i]){ 41 prime[++tot]=i,low[i]=i,d[i]=2,sqr[i]=(i*i+1)%mod; 42 for(int j=i*i;j<=n;j*=i) d[j]=d[j/i]+1,sqr[j]=(sqr[j/i]+j*j)%mod; 43 } 44 for(int j=1;j<=tot && i*prime[j]<=n;j++){ 45 is[i*prime[j]]=0; 46 if(i%prime[j]==0){ 47 low[i*prime[j]]=low[i]*prime[j]; 48 if(low[i]!=i){ 49 d[i*prime[j]]=d[i/low[i]]*d[low[i]*prime[j]]; 50 sqr[i*prime[j]]=sqr[i/low[i]]*sqr[low[i]*prime[j]]%mod; 51 } 52 break; 53 } 54 low[i*prime[j]]=prime[j]; 55 d[i*prime[j]]=d[i]*d[prime[j]]; 56 sqr[i*prime[j]]=sqr[i]*sqr[prime[j]]%mod; 57 } 58 } 59 } 60 61 inline int main(){ 62 q=read(),qi=read(),A=read(),B=read(),C=read(); 63 init(C); 64 while(q--){ 65 ans1=(ans1+d[qi]+(qi&1))%mod; 66 ans2=(ans2+sqr[qi]+(qi&1)*4)%mod; 67 qi=(qi*A+B)%C+1; 68 } 69 printf("%lld\n%lld\n",ans1,ans2); 70 return 0; 71 } 72 } 73 signed main(){return CL::main();}
附加题：中国象棋
比较好的DP题
考虑根据定义，同一行同一列不能同时存在三个炮，并且题目要求为计数
那么计数题通常由于方案数极多并不能直接考虑每种方案，需要忽略一些信息
来降低问题复杂度，对于本题而言，只需要考虑棋子的放置方案，对于棋子
放置的具体位置并不关心，于是考虑设计f[i][j][k]表示只考虑前i行其中第i行
有j列放置1个棋子，k列放置2个棋子，这样既保证转移的基本需要，有避开
了棋子位置，转移时枚举上一次决策大力分类讨论即可，即不放棋子：直接继承
放置一个棋子：考虑放置的位置，放置两个棋子：考虑分别放置在哪即可
代码如下：
1 %: pragma GCC optimize(3) 2 #include <bits/stdc++.h> 3 using namespace std; 4 #define I long long 5 #define C char 6 #define B bool 7 #define V void 8 #define D double 9 #define LL long long 10 #define UI unsigned int 11 #define UL unsigned long long 12 #define P pair<I,I> 13 #define MP make_pair 14 #define a first 15 #define b second 16 #define lowbit(x) (x & -x) 17 #define debug cout << "It's Ok Here !" << endl; 18 #define FP(x) freopen (#x,"r",stdin) 19 #define FC(x) freopen (#x,"w",stdout) 20 const I N = 105, mod = 9999973; 21 I n,m,ans,f[N][N][N]; 22 inline I read () { 23 I x(0),y(1); C z(getchar()); 24 while (!isdigit(z)) { if (z == '-') y = -1; z = getchar(); } 25 while ( isdigit(z)) x = x * 10 + (z ^ 48), z = getchar(); 26 return x * y; 27 } 28 inline V Max (I &a,I b) { a = a > b ? a : b; } 29 inline V Min (I &a,I b) { a = a < b ? a : b; } 30 inline I max (I a,I b) { return a > b ? a : b; } 31 inline I min (I a,I b) { return a < b ? a : b; } 32 inline V swap (I &a,I &b) { a ^= b, b ^= a, a ^= b; } 33 inline I abs (I a) { return a >= 0 ? a : -a; } 34 inline P operator + (const P &a,const P &b) { 35 return MP (a.a + b.a,a.b + b.b); 36 } 37 inline P operator - (const P &a,const P &b) { 38 return MP (a.a - b.a,a.b - b.b); 39 } 40 inline V Mod1 (I &a,I b) { a = a + b > mod ? a + b - mod : a + b; } 41 signed main () { 42 FP (chess.in), FC (chess.out); 43 n = read (), m = read (); 44 f[0][0][0] = 1; 45 for (I i(1);i <= n; ++ i) { 46 for (I j(0);j <= m; ++ j) { 47 for (I k(0);j + k <= m; ++ k) { 48 f[i][j][k] = f[i - 1][j][k]; 49 if (k >= 1) f[i][j][k] = (f[i][j][k] + f[i - 1][j + 1][k - 1] * (j + 1)); 50 if (j >= 1) f[i][j][k] = (f[i][j][k] + f[i - 1][j - 1][k] * (m - j + 1 - k)); 51 if (k >= 1) f[i][j][k] = (f[i][j][k] + f[i - 1][j][k - 1] * (m - k + 1 - j) * j); 52 if (k >= 2) f[i][j][k] = (f[i][j][k] + f[i - 1][j + 2][k - 2] * (j + 2) * (j + 1) / 2); 53 if (j >= 2) f[i][j][k] = (f[i][j][k] + f[i - 1][j - 2][k] * (m - j + 2 - k) * (m - j + 1 - k) / 2); 54 f[i][j][k] %= mod; if (i == n) Mod1 (ans,f[i][j][k]); 55 } 56 } 57 } 58 printf ("%lld\n",ans); 59 }
