A: 签到. 我 wa 了一发怎么办啊.
- #include<bits/stdc++.h>
- using namespace std;
- #define ll long long
- #define inf 1000000010
- 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;
- }
- signed main()
- {
- int n=read();
- cout<<(n-2)*180;
- return 0;
- //NOTICE LONG LONG!!!!!
- }
B: 签到.
- #include<bits/stdc++.h>
- using namespace std;
- #define ll long long
- #define inf 1000000010
- 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;
- }
- char s[20];
- signed main()
- {
- scanf("%s",s+1);
- int cnt=0;
- for (int i=1;i<=strlen(s+1);i++) if (s[i]=='x') cnt++;
- if (cnt>=8) cout<<"NO";else cout<<"YES";
- return 0;
- //NOTICE LONG LONG!!!!!
- }
C: 考虑枚举最后两人各胜多少局. 注意到期望每 100/(100-C)局就会决出一次胜负, 于是只需要考虑该种胜负局数出现概率. 不妨设第一个人赢了 n 场输了 x 场, 那么概率就是 C(n+x-1,n-1).An.Bx/(A+B)n+x, 累加即可.
- #include<bits/stdc++.h>
- using namespace std;
- #define ll long long
- #define inf 1000000010
- #define N 100010
- #define P 1000000007
- 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,A,B,C,f[1000][1000],fac[N<<1],Inv[N<<1],ans;
- int ksm(int a,int k)
- {
- int s=1;
- for (;k;k>>=1,a=1ll*a*a%P) if (k&1) s=1ll*s*a%P;
- return s;
- }
- int inv(int a){return ksm(a,P-2);}
- void inc(int &x,int y){x+=y;if (x>=P) x-=P;}
- int calc(int n,int m){if (m>n) return 0;return 1ll*fac[n]*Inv[m]%P*Inv[n-m]%P;}
- signed main()
- {
- n=read(),A=read(),B=read(),C=read();A=1ll*A*inv(100-C)%P,B=1ll*B*inv(100-C)%P;
- fac[0]=1;for (int i=1;i<=2*n;i++) fac[i]=1ll*fac[i-1]*i%P;
- Inv[0]=Inv[1]=1;for (int i=2;i<=2*n;i++) Inv[i]=P-1ll*(P/i)*Inv[P%i]%P;
- for (int i=2;i<=2*n;i++) Inv[i]=1ll*Inv[i]*Inv[i-1]%P;
- for (int i=0;i<n;i++)
- {
- int p1=1ll*A%P*calc(n+i-1,n-1)%P*ksm(A,n-1)%P*ksm(B,i)%P;
- int p2=1ll*B%P*calc(n+i-1,n-1)%P*ksm(B,n-1)%P*ksm(A,i)%P;
- inc(ans,1ll*(n+i)*100%P*inv(100-C)%P*(p1+p2)%P);
- }
- cout<<ans;
- return 0;
- //NOTICE LONG LONG!!!!!
- }
D: 大胆猜想从小到大填每次选 (未被占用) 度数最小的点即可.
- #include<bits/stdc++.h>
- using namespace std;
- #define ll long long
- #define inf 1000000010
- #define N 10010
- 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,p[N],a[N],degree[N],val[N],t,ans;
- struct data{int to,nxt;
- }edge[N<<1];
- void addedge(int x,int y){t++;edge[t].to=y,edge[t].nxt=p[x],p[x]=t;}
- signed main()
- {
- n=read();
- for (int i=1;i<n;i++)
- {
- int x=read(),y=read();
- addedge(x,y),addedge(y,x);
- degree[x]++,degree[y]++;
- }
- for (int i=1;i<=n;i++) a[i]=read();
- sort(a+1,a+n+1);
- for (int i=1;i<=n;i++)
- {
- int mn=n;
- for (int j=1;j<=n;j++)
- if (val[j]==0) mn=min(mn,degree[j]);
- for (int j=1;j<=n;j++)
- if (val[j]==0&°ree[j]==mn) {mn=j;break;}
- val[mn]=a[i];ans+=degree[mn]*a[i];
- for (int j=p[mn];j;j=edge[j].nxt)
- degree[edge[j].to]--;
- }
- cout<<ans<<endl;
- for (int i=1;i<=n;i++) cout<<val[i]<<' ';
- return 0;
- //NOTICE LONG LONG!!!!!
- }
E: 每一项都除掉 d. 特判 d=0. 无所事事了 1h 后在最后 20s 想到了做法.
- #include<bits/stdc++.h>
- using namespace std;
- #define ll long long
- #define inf 1000000010
- #define P 1000003
- 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 Q,fac[P],inv[P];
- int ksm(int a,int k)
- {
- int s=1;
- for (;k;k>>=1,a=1ll*a*a%P) if (k&1) s=1ll*s*a%P;
- return s;
- }
- int I(int a){return ksm(a,P-2);}
- signed main()
- {
- Q=read();
- fac[0]=1;for (int i=1;i<P;i++) fac[i]=1ll*fac[i-1]*i%P;
- inv[0]=inv[1]=1;for (int i=2;i<P;i++) inv[i]=P-1ll*(P/i)*inv[P%i]%P;
- for (int i=2;i<P;i++) inv[i]=1ll*inv[i-1]*inv[i]%P;
- while (Q--)
- {
- int x=read(),d=read(),n=read();
- if (d==0) {printf("%d\n",ksm(x,n));continue;}
- x=1ll*x*I(d)%P;
- if (x+n-1>=P) printf("%d\n",0);
- else printf("%d\n",1ll*fac[x+n-1]*(x==0?1:inv[x-1])%P*ksm(d,n)%P);
- }
- return 0;
- //NOTICE LONG LONG!!!!!
- }
- result:rank 191 rating +9
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