20160813 PM Training Record

* In the afternoon I heard the lecture was simple, and was pulled to the next door to do a tasty hu not to write the question

T1

Jiangsu Province when the choice of a topic, Shao crazy at that time with my mouth hu over this problem

Let's just not think about 1.

Then we select an element, when the second one is added, the parity of the two is different, when the third number is added, it is definitely not satisfied.

So we're going to have to give you a 1 count, and then we'll mess up.

intn,a[1234];BOOLnp[33333333];inttot,pr[8001234];Const intn=30000000;voidsieve () { for(intI=2; i<=n;i++){        if(!np[i]) pr[++tot]=i;  for(intj=1; j<=tot&&i*pr[j]<=n;j++) {Np[i*pr[j]]=1; if(i%pr[j]==0) Break; }    }    return;}intFintcnt1) {    intans=0; if(CNT1) for(intI=1; i<=n;i++)if(a[i]!=1)if(!np[a[i]+1])return 1; Else         return 0;}intMain () {Freopen ("prime.in","R", stdin); Freopen ("Prime.out","W", stdout);    Sieve (); CIN>>N; intL=0, cnt1=0, is2=0;  for(intI=1; i<=n;i++) cin>>a[i],cnt1+= (a[i]==1); Sort (a+1, a+n+1); if(n==2) {cout<<n<<endl<<a[1]<<' '<<a[2]; return 0; }     for(intI=1; i<=n;i++) for(intj=1; j<=n;j++)if(I^J)if(!np[a[i]+a[j]]) L=2; if(! L for(intI=1; i<=n;i++)if(!np[a[i]]) L=1; if(l==1) {printf ("1\n"); for(inti=n;i>=1; i--)if(!Np[a[i]]) {printf ("%d", A[i]);return 0; }    }intT; if(Cnt1+f (CNT1) <=2) {puts ("2");  for(inti=n;i>=1; i--) for(intj=i-1; j>=1; j--){        if(I^J)if(!np[a[i]+a[j]]) printf ("%d%d\n", A[j],a[i]);return 0;} }Else{L=cnt1+f (cnt1); cout<<L<<Endl;  for(intI=1; i<=cnt1;i++) printf ("%d",1);  for(inti=n;i>=1; i--)if(!np[a[i]+1]) {printf ("%d", A[i]);return 0; }    }}

Then I went away.

T2

will only N^2DP

F[I][J][0] The optimal value of the first I put J Flower, I do not put

F[I][J][1] The optimal value of the first I put J Flower, I put

Transfer

intf[3010][3010][2],a[3010];intn,m;intMain () {Freopen ("flower.in","R", stdin); Freopen ("Flower.out","W", stdout); scanf ("%d%d",&n,&m);  for(intI=1; i<=n;i++) scanf ("%d",&A[i]);  for(intI=1; i<=n;i++) for(intj=1; j<=m;j++) f[i][j][0]=f[i][j][1]=-2000000000; f[1][1][1]=a[1];  for(intI=2; i<=n;i++) {f[i][0][0]=0;  for(intj=1; j<=m;j++) {f[i][j][1]=f[i-1][j-1][0]+A[i]; f[i][j][0]=max (f[i-1][j][0],f[i-1][j][1]); }    }    intans=f[n][m][0];  for(intI=1; i<=n;i++) for(intj=1; j<=m;j++) f[i][j][0]=f[i][j][1]=-2000000000; f[1][0][0]=0;  for(intI=2; i<=n;i++) {f[i][0][0]=0;  for(intj=1; j<=m;j++) {f[i][j][1]=f[i-1][j-1][0]+A[i]; f[i][j][0]=max (f[i-1][j][0],f[i-1][j][1]); }} ans=max (Ans,max (f[n][m][0],f[n][m][1])); if(m>n/2) cout<<"error!"; Elsecout<<ans;//cout<< (M>N/2? ") Error! ": ans);    return 0;}

T3

Given a tree, you can delete an edge, divide it into two unicom blocks, and minimize the sum of two diameters.

Won't do

20160813 PM Training Record