A：n==2?2:1。
#include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<algorithm> using namespace std; #define ll long long char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;} int gcd(int n,int m){return m==0?n:gcd(m,n%m);} int read() { int x=0,f=1;char c=getchar(); while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();} while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar(); return x*f; } int n; int main() { /*#ifndef ONLINE_JUDGE freopen("a.in","r",stdin); freopen("a.out","w",stdout); #endif*/ n=read(); if (n==2) cout<<2;else cout<<1; return 0; }
B：sort一下，组内人数相同的放一组，超了就换一组。
#include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<algorithm> using namespace std; #define ll long long #define N 100010 char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;} int gcd(int n,int m){return m==0?n:gcd(m,n%m);} int read() { int x=0,f=1;char c=getchar(); while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();} while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar(); return x*f; } int n,ans[N],cnt; struct data { int x,y; bool operator <(const data&a) const { return x<a.x; } }a[N]; int main() { /*#ifndef ONLINE_JUDGE freopen("a.in","r",stdin); freopen("a.out","w",stdout); #endif*/ n=read(); for (int i=1;i<=n;i++) a[i].x=n-read(),a[i].y=i; sort(a+1,a+n+1); for (int i=1;i<=n;i++) { int t=i; while (t<n&&a[t+1].x==a[i].x&&t+1-i+1<=a[i].x) t++; if (t-i+1!=a[i].x) {cout<<"Impossible";return 0;} cnt++; for (int j=i;j<=t;j++) ans[a[j].y]=cnt; i=t; } printf("Possible\n"); for (int i=1;i<=n;i++) printf("%d ",ans[i]); return 0; }
C：f[i][j]前i个有j块那啥的方案数，转移显然。
#include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<algorithm> using namespace std; #define ll long long #define P 998244353 #define N 2010 char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;} int gcd(int n,int m){return m==0?n:gcd(m,n%m);} int read() { int x=0,f=1;char c=getchar(); while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();} while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar(); return x*f; } int n,m,k,f[N][N]; int main() { /*#ifndef ONLINE_JUDGE freopen("a.in","r",stdin); freopen("a.out","w",stdout); #endif*/ n=read(),m=read(),k=read(); f[1][0]=m; for (int i=2;i<=n;i++) { f[i][0]=m; for (int j=1;j<=k;j++) f[i][j]=(f[i-1][j]+1ll*f[i-1][j-1]*(m-1))%P; } cout<<f[n][k]; return 0; }
D：kruskal求出最小瓶颈树，最后使所有标记点连接成一个连通块的边即为答案。莫名其妙的写了一堆东西还wa了两发。
#include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<algorithm> using namespace std; #define ll long long #define N 200010 char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;} int gcd(int n,int m){return m==0?n:gcd(m,n%m);} int read() { int x=0,f=1;char c=getchar(); while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();} while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar(); return x*f; } int n,m,k,fa[N]; bool flag[N],f[N]; struct data { int x,y,z; bool operator <(const data&a) const { return z<a.z; } }e[N]; namespace tree { int p[N],t,size[N],dfn[N],cnt; struct data{int to,nxt,len; }edge[N<<1]; void addedge(int x,int y,int z){t++;edge[t].to=y,edge[t].nxt=p[x],edge[t].len=z,p[x]=t;} void dfs(int k,int from) { size[k]=f[k];dfn[k]=++cnt; for (int i=p[k];i;i=edge[i].nxt) if (edge[i].to!=from) { dfs(edge[i].to,k); size[k]+=size[edge[i].to]; } } } int find(int x){return fa[x]==x?x:fa[x]=find(fa[x]);} int main() { /*#ifndef ONLINE_JUDGE freopen("a.in","r",stdin); freopen("a.out","w",stdout); #endif*/ n=read(),m=read(),k=read(); for (int i=1;i<=k;i++) f[read()]=1; /*for (int i=1;i<=m;i++) { int x=read(),y=read(),z=read(); addedge(x,y,z),addedge(y,x,z); }*/ for (int i=1;i<=m;i++) e[i].x=read(),e[i].y=read(),e[i].z=read(); sort(e+1,e+m+1); for (int i=1;i<=n;i++) fa[i]=i; for (int i=1;i<=m;i++) { int p=find(e[i].x),q=find(e[i].y); if (p!=q) fa[q]=p,tree::addedge(e[i].x,e[i].y,e[i].z),tree::addedge(e[i].y,e[i].x,e[i].z),flag[i]=1; } tree::dfs(1,1); for (int i=m;i>=1;i--) if (flag[i]) { int x=e[i].x,y=e[i].y; if (tree::dfn[x]<tree::dfn[y]) swap(x,y); if (tree::size[x]&&tree::size[x]<k) {for (int j=1;j<=k;j++) printf("%d ",e[i].z);return 0;} } return 0; }
E：真的自闭了。本身就非常思博还想了50min。肝F无果回来hack的时候，在最后1min发现了输出没开I64d，居然还犹豫了一下是不是实际上并不会出问题，然后还真的就没改。看来就算define int long long也解决不了我天天爆int了。
注意到对于偶数位我们应该让这里的前缀和尽量小，因为下一位可以任意取。于是记录到每个偶数位时的最小前缀和即可。同时由平方差公式，可以暴力枚举偶数位上的数的因数来快速枚举每种情况。
#include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<algorithm> #include<vector> using namespace std; #define ll long long #define N 200010 char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;} int gcd(int n,int m){return m==0?n:gcd(m,n%m);} int read() { int x=0,f=1;char c=getchar(); while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();} while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar(); return x*f; } int n,f[N]; const double eps=1E-9; ll a[N]; int main() { /*#ifndef ONLINE_JUDGE freopen("a.in","r",stdin); freopen("a.out","w",stdout); #endif*/ n=read(); for (int i=1;i<=n/2;i++) a[i<<1]=read(); for (int i=2;i<=n;i+=2) { for (int j=1;j*j<a[i];j++) if (a[i]%j==0) { int x=j,y=a[i]/j;//q-p=x q+p=y if ((x+y&1)||(y-x&1)) continue; int q=x+y>>1,p=y-x>>1;//p i-1ǰ q iǰ if (f[i-2]<p) f[i]=q,a[i-1]=1ll*p*p-1ll*f[i-2]*f[i-2]; } if (f[i]==0) {cout<<"No";return 0;} } cout<<"Yes\n"; for (int i=1;i<=n;i++) printf("%I64d ",a[i]); return 0; }
一夜回到解放前。虽然看起来大号还是不会变成小号的。
不过好像D和E都fst了不少？
然后我还发现F原来是真随机本来一直在想靠谱做法。
再也不打Chinese Round了