Magic W is small total number of program subtracts the number of satisfactory solution considering the original sequence s sequence ends put together, or just one of two ends of look at both ends of the split program, to determine the sequence of prefix length x S is the same prefix, then there is a suffix length len-x S with a suffix of the same order do not leak, x + 1 position sequence is not necessarily equal to X 1 + S position, since in this case the length is x + prefixed same S case 1 will be taken into account, the total number of programs $ 25 * 26 ^ {n- len-1} $ enumeration x find x to len (prefix is equal to S is also taken into account), then program number $ len * 25 * 26 ^ { n-len-1} $ last seen suffix S solutions, apparently finally len is determined $ 26 ^ {n-len} $, but also because when the sequence first letter equal to S and a first suffix letter S and the same situation will be considered in x == 1 went in so determined first letter of the sequence, there are 25 cases, the number of programs that suffix $ 25 * 26 ^ {n- len} $ The total program $ 26 ^ n-25 * 26 ^ {n-len-1} * len-25 * 26 ^ {n-len} $
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 using namespace std; 5 long long n,len; 6 char s[1100000]; 7 const long long mod=998244353; 8 long long qpow(long long a,long long b){ 9 if(b<0) return 0; 10 long long ans=1; 11 a%=mod; 12 while(b){ 13 if(b&1) ans=ans*a%mod; 14 b>>=1; 15 a=a*a%mod; 16 } 17 return ans; 18 } 19 int main(){ 20 freopen("magic.in","r",stdin); 21 freopen("magic.out","w",stdout); 22 scanf("%lld%s",&n,s+1); 23 len=strlen(s+1); 24 long long ans=0; 25 ans=(ans+qpow(26,n-len-1)*25%mod*len%mod)%mod; 26 ans=(ans+qpow(26,n-len))%mod; 27 printf("%lld\n",(qpow(26,n)-ans+mod)%mod); 28 }
FIG small Y
trucking original title, maintaining the minimum spanning tree path with the maximum multiplier tree +
1 #include<iostream> 2 #include<cstdio> 3 #include<algorithm> 4 using namespace std; 5 struct node{ 6 int l,r,z; 7 bool operator < (const node b)const{ 8 return z<b.z; 9 } 10 }e[310000]; 11 int n,m,Q,ances[310000],fa[310000][20],maxn[310000][20],dis[310000]; 12 int to[610000],nex[610000],head[310000],c[610000],tot; 13 int find(int x){return ances[x]==x?x:ances[x]=find(ances[x]);} 14 void add(int x,int y,int z){ 15 to[++tot]=y,nex[tot]=head[x],head[x]=tot,c[tot]=z; 16 } 17 void dfs(int x,int pre){ 18 dis[x]=dis[pre]+1; 19 fa[x][0]=pre; 20 for(register int i=1;i<=18;i++){ 21 fa[x][i]=fa[fa[x][i-1]][i-1]; 22 maxn[x][i]=max(maxn[x][i-1],maxn[fa[x][i-1]][i-1]); 23 } 24 for(register int i=head[x];i;i=nex[i]){ 25 int y=to[i]; 26 if(y==pre) continue; 27 maxn[y][0]=c[i]; 28 dfs(y,x); 29 } 30 } 31 int lca(int x,int y){ 32 if(dis[x]>dis[y]) swap(x,y); 33 int ans=0; 34 for(register int i=18;i>=0;i--){ 35 if(dis[fa[y][i]]>=dis[x]){ 36 ans=max(ans,maxn[y][i]); 37 y=fa[y][i]; 38 } 39 } 40 if(x==y) return ans; 41 for(register int i=18;i>=0;i--){ 42 if(fa[y][i]!=fa[x][i]){ 43 ans=max(ans,maxn[y][i]); 44 ans=max(ans,maxn[x][i]); 45 y=fa[y][i]; 46 x=fa[x][i]; 47 } 48 } 49 return max(ans,max(maxn[x][0],maxn[y][0])); 50 } 51 int main(){ 52 freopen("graph.in","r",stdin); 53 freopen("graph.out","w",stdout); 54 scanf("%d%d%d",&n,&m,&Q); 55 for(register int i=1;i<=m;i++) scanf("%d%d%d",&e[i].l,&e[i].r,&e[i].z); 56 sort(e+1,e+m+1); 57 for(register int i=1;i<=n;i++) ances[i]=i; 58 for(register int i=1;i<=m;i++){ 59 int x=find(e[i].l),y=find(e[i].r); 60 if(x!=y){ 61 add(e[i].l,e[i].r,e[i].z); 62 add(e[i].r,e[i].l,e[i].z); 63 ances[x]=y; 64 } 65 } 66 for(register int i=1;i<=n;i++){ 67 if(!dis[i]){ 68 dfs(i,0); 69 } 70 } 71 int x,y; 72 while(Q--){ 73 scanf("%d%d",&x,&y); 74 if(find(x)!=find(y)) puts("-1"); 75 else printf("%d\n",lca(x,y)); 76 } 77 }
The number of small L
The answer must be less than or equal to 4, because any number can be composed of 01,02,04,08 ((0 to 9) each, can be composed of 0,1,2,4,8)
Verification, two, three favorite numbers can spell, otherwise 4
From low to high numbers in the query enumerator digits, three-dimensional DP f [i] [j] [k], represents the i-th bit down into a 0 or 1 or 2, and there is k (k <8) state indicates whether or not each number to the highest three bits of the program whether there
Simple DP, and finally asked to see the highest bit of the presence or absence can be 0
Online slower seek time, vigorously card often enough, large pre-rolled my God yxs
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #define mu(x) ((x)<10?(x):((x)>=20?(x)-20:(x)-10)) 5 #define div(x) ((x)<10?0:((x)>=20?2:1)) 6 using namespace std; 7 int T,a,b,c,d,u,v; 8 char f[20][3][1<<3],p[20]; 9 long long now; 10 bool judge(){ 11 memset(f,0,sizeof(f)); 12 f[0][0][7]=1; 13 for(int t=1;t<=p[0];t++){ 14 register int w=p[t],flag=0; 15 for(register int j=0;j<=2;j++){ 16 if(f[t-1][j][1]||f[t-1][j][3]||f[t-1][j][5]||f[t-1][j][7]){ 17 if(mu(a+j)==w) f[t][div(a+j)][1]=1; 18 if(mu(b+j)==w) f[t][div(b+j)][1]=1; 19 } 20 if(f[t-1][j][2]||f[t-1][j][3]||f[t-1][j][6]||f[t-1][j][7]){ 21 if(mu(c+j)==w) f[t][div(c+j)][2]=1; 22 if(mu(d+j)==w) f[t][div(d+j)][2]=1; 23 } 24 if(f[t-1][j][4]||f[t-1][j][5]||f[t-1][j][6]||f[t-1][j][7]){ 25 if(mu(u+j)==w) f[t][div(u+j)][4]=1; 26 if(mu(v+j)==w) f[t][div(v+j)][4]=1; 27 } 28 if(f[t-1][j][3]||f[t-1][j][7]){ 29 if(mu(a+c+j)==w) f[t][div(a+c+j)][3]=1; 30 if(mu(a+d+j)==w) f[t][div(a+d+j)][3]=1; 31 if(mu(b+c+j)==w) f[t][div(b+c+j)][3]=1; 32 if(mu(b+d+j)==w) f[t][div(b+d+j)][3]=1; 33 } 34 if(f[t-1][j][5]||f[t-1][j][7]){ 35 if(mu(a+u+j)==w) f[t][div(a+u+j)][5]=1; 36 if(mu(a+v+j)==w) f[t][div(a+v+j)][5]=1; 37 if(mu(b+u+j)==w) f[t][div(b+u+j)][5]=1; 38 if(mu(b+v+j)==w) f[t][div(b+v+j)][5]=1; 39 } 40 if(f[t-1][j][6]||f[t-1][j][7]){ 41 if(mu(c+u+j)==w) f[t][div(c+u+j)][6]=1; 42 if(mu(c+v+j)==w) f[t][div(c+v+j)][6]=1; 43 if(mu(d+u+j)==w) f[t][div(d+u+j)][6]=1; 44 if(mu(d+v+j)==w) f[t][div(d+v+j)][6]=1; 45 } 46 if(f[t-1][j][7]){ 47 if(mu(a+c+u+j)==w) f[t][div(a+c+u+j)][7]=1; 48 if(mu(a+c+v+j)==w) f[t][div(a+c+v+j)][7]=1; 49 if(mu(a+d+u+j)==w) f[t][div(a+d+u+j)][7]=1; 50 if(mu(a+d+v+j)==w) f[t][div(a+d+v+j)][7]=1; 51 if(mu(b+c+u+j)==w) f[t][div(b+c+u+j)][7]=1; 52 if(mu(b+c+v+j)==w) f[t][div(b+c+v+j)][7]=1; 53 if(mu(b+d+u+j)==w) f[t][div(b+d+u+j)][7]=1; 54 if(mu(b+d+v+j)==w) f[t][div(b+d+v+j)][7]=1; 55 } 56 } 57 if(!(f[t][0][1]+f[t][0][2]+f[t][0][3]+f[t][0][4]+f[t][0][5]+f[t][0][6]+f[t][0][7]+f[t][1][1]+f[t][1][2]+f[t][1][3]+f[t][1][4]+f[t][1][5]+f[t][1][6]+f[t][1][7]+f[t][2][1]+f[t][2][2]+f[t][2][3]+f[t][2][4]+f[t][2][5]+f[t][2][6]+f[t][2][7])) return 0; 58 } 59 for(register int i=1;i<=7;i++) if(f[p[0]][0][i]) return 1; 60 return 0; 61 } 62 int work(){ 63 p[0]=0; 64 for(long long i=now;i;i/=10) p[++p[0]]=i%10; 65 for(a=0;a<=9;a++) for(b=a+(a!=0);b<=9;b++) 66 for(c=a;c<=9;c++) for(d=c+(c!=0);d<=9;d++) 67 for(u=c;u<=9;u++) for(v=u+(u!=0);v<=9;v++) 68 if(judge()){ 69 int ans=0; 70 if(a||b) ans++; 71 if(c||d) ans++; 72 if(u||v) ans++; 73 return ans; 74 } 75 return 0; 76 } 77 int main(){ 78 freopen("number.in","r",stdin); 79 freopen("number.out","w",stdout); 80 scanf("%d",&T); 81 while(T--){ 82 scanf("%lld",&now); 83 int ans=work(); 84 if(ans) printf("%d\n",years); 85 else puts("4"); 86 } 87 }