代码不想写就口胡了
1001.
用 歌唱王国 的结论
$E(A)=\sum_{i=1}^{len} a_{i}*m^i,\ a_{i}=is\_border(i)$
回文串border就是回文自动机的fail
直接上回文自动机比较字典序即可
1002.
1003.
直接离线考虑每一位的贡献
用个dfs序维护线段 树去掉u,v是祖先儿子的情况就行
$O(20*n*logn)$
1004.
$sum_{i}=(sum_{i-1}+a[i])\%mod$
$f_{i}=max(f_{i-1},max\{f_{j}|sum(j)=sum(i)\}+1)$
1005.
没看
1006.
数位dp
写了一个小时，真实自闭
#include <bits/stdc++.h> #define rep(i,a,b) for(int i=a;i<=b;++i) #define per(i,a,b) for(int i=a;i>=b;--i) using namespace std; typedef long long s64; inline int read() { int f=1,sum=0; char x=getchar(); for(;(x<'0'||x>'9');x=getchar()) if(x=='-') f=-1; for(;x>='0'&&x<='9';x=getchar()) sum=sum*10+x-'0'; return f*sum; } int len; int wei[20]; int cnt[10]; void divid_(s64 n) { len=0; while (n) { wei[++len]=n%10; n/=10; } } s64 f[2][20]; s64 C[20][20]; s64 get_C(int x,int y) { return C[x][y]; } s64 dp(int x,int d,bool is_zero) { s64 ans=0; if(d==0) { // cout<<"-_-x:"<<x<<endl; rep(i,0,x) { // i=2; memset(f,0,sizeof(f)); int opt=0; f[0][i]=get_C(x-is_zero,i); //cout<<i<<" "<<f[0][i]<<endl; rep(j,1,9) { opt^=1; memset(f[opt],0,sizeof(f[opt])); //cout<<j<<" "<<i+cnt[d]-1-cnt[j]<<" "<<x-cnt[j]<<endl; rep(u,0,i+cnt[d]-1-cnt[j]) rep(k,0,x-u) f[opt][u+k]+=f[opt^1][k]*get_C(x-k,u); } ans+=f[opt][x]; //cout<<"ans:"<<ans<<endl; //while (1); } } else { rep(i,0,x) { memset(f,0,sizeof(f)); int opt=0; rep(j,0,i+cnt[d]-1-cnt[0]) f[0][j]=get_C(x-is_zero,j); rep(j,1,9) { if(d==j) { opt^=1; memset(f[opt],0,sizeof(f[opt])); rep(k,i,x) f[opt][k]=f[opt^1][k-i]*get_C(x-k+i,i); continue; } opt^=1; memset(f[opt],0,sizeof(f[opt])); rep(u,0,i+cnt[d]-1-cnt[j]) rep(k,0,x-u) f[opt][u+k]+=f[opt^1][k]*get_C(x-k,u); } ans+=f[opt][x]; } } return ans; } bool check(int d) { int aa[10]={0}; rep(i,1,len) ++aa[wei[i]]; rep(i,0,9) if(i!=d&&aa[i]>=aa[d]) return 0; return 1; } s64 calc(s64 n,int d) { divid_(n); memset(cnt,0,sizeof(cnt)); s64 ans=0; rep(i,1,len-1) ans+=dp(i,d,1); per(i,len,1) { //cout<<"-_-"<<ans<<endl; rep(j,i==len,wei[i]-1) { ++cnt[j]; ans+=dp(i-1,d,0); //cout<<"-_-"<<i<<" "<<j<<" "<<dp(i-1,d,0)<<endl; --cnt[j]; } ++cnt[wei[i]]; } if(check(d)) ++ans; return ans; } signed main () { //freopen("a.txt","r",stdin); //freopen("a.out","w",stdout); int test_=read(); rep(i,0,19) C[i][0]=1; rep(i,1,19) rep(j,1,19) C[i][j]=C[i-1][j]+C[i-1][j-1]; while (test_--) { s64 l,r; int d; cin>>l>>r>>d; //cout<<calc(l-1,d)<<endl; // cout<<calc(l-1,d)<<" "<<calc(r,d)<<endl; cout<<calc(r,d)-calc(l-1,d)<<endl; } }
1007.
脑残题，不知道为什么边权是随机的
搞了半天自己取的是min还能过样例
性质就是最短路径很短，然后直接搜就行了
#include <bits/stdc++.h> #define rep(i,a,b) for(int i=a;i<=b;++i) #define per(i,a,b) for(int i=a;i>=b;--i) using namespace std; typedef long long s64; inline int read() { int f=1,sum=0; char x=getchar(); for(;(x<'0'||x>'9');x=getchar()) if(x=='-') f=-1; for(;x>='0'&&x<='9';x=getchar()) sum=sum*10+x-'0'; return f*sum; } const int M=1e2+5; int n,m; int a[M][M]; int ans; bool can[M][M][M]; bool vis[M][M]; int dis[M],pre[M]; bool inq_[M]; void floyd(int x) { memset(can[x],0,sizeof(can[x])); memset(dis,0x3f,sizeof(dis)); dis[1]=0; queue<int> q; q.push(1); while (!q.empty()) { int r=q.front(); q.pop(); inq_[r]=0; rep(i,1,n) if(!vis[r][i]&&dis[i]>dis[r]+a[r][i]) { dis[i]=dis[r]+a[r][i]; pre[i]=r; if(!inq_[i]) inq_[i]=1,q.push(i); } } int now=n; while (now!=1) { can[x][pre[now]][now]=can[x][now][pre[now]]=1; now=pre[now]; } } void calc() { floyd(10); ans=max(ans,dis[n]); } void dfs(int x) { if(x==m+1) return calc(),void(); floyd(x); rep(i,1,n) rep(j,i+1,n) if(can[x][i][j]) { vis[i][j]=vis[j][i]=1; dfs(x+1); vis[i][j]=vis[j][i]=0; } } signed main () { //freopen("a.txt","r",stdin); //freopen("a.out","w",stdout); int test_=read(); while (test_--) { n=read(),m=read(); //cout<<n<<" "<<m<<endl; rep(i,1,n) rep(j,i+1,n) { int x=read(),y=read(); a[x][y]=a[y][x]=read(); } ans=0; memset(can,0,sizeof(can)); dfs(1); cout<<ans<<endl; } }
1008.
关键就是把反射问题变成无线个三角形放在平面上的情况
二分答案，判断能相交多少次就行，直接用长度除以间距即可
1009.
好像比较简单，没看
1010.
1011.