Codeforces #298题解

Question A

There are a lot of feasible construction methods that I am taking for each number I and I+N/2 pairing way. For 4 special treatment, for 2 4 1 3.

#include <iostream>#include<string>using namespacestd;intMain () {intN; CIN>>N; if(n==1) {cout<<1<<Endl; cout<<1<<Endl; }    Else if(n==2) {cout<<1<<Endl; cout<<1<<Endl; }        Else if(n==3) {cout<<2<<Endl; cout<<"1 3"<<Endl; }    Else if(n==4) {cout<<4<<Endl; cout<<"2 4 1 3"<<Endl; }    Else    {        if(n%2==0) {cout<<n<<Endl; cout<<1<<" "<<1+n/2;  for(intI=2; i<=n/2; i++) cout<<" "<<i<<" "<<i+n/2; cout<<Endl; }        Else{cout<<n<<Endl;  for(intI=1;i< (n+1)/2; i++) cout<<i<<" "<<i+ (n+1)/2<<" "; cout<< (n+1)/2; cout<<Endl; }    }}

B: For each d>=i>=-d enumeration, fetch can satisfy the maximum value that can be reached in the remaining time v2. That is the greedy law.

#include <iostream>#include<cstdio>#include<algorithm>using namespacestd;intV1,v2;intt,d,sum;intMain () {CIN>>v1>>v2>>t>>D; if(V2 <v1) swap (V1,V2); Sum=v1;  for(intI=2; i<=t;i++)    {         for(intj=d;j>=-d;j--)        {            if(v1+j-(t-i) *d<=v2) {V1+=J;  Break; }} sum+=v1; } cout<<sum<<Endl; return 0;}

C: First to find the upper limit (except yourself, the other is at least 1 o'clock) Max, because the data is 10^5, so pre-processing out sum[i], the first I and. This can be calculated by tot and sum[when each lower limit is taken.

Note the situation of the lower limit temp<0 special treatment.

#include <iostream>#include<cstring>#include<string>#include<cmath>#include<algorithm>using namespacestd;Long Longsum[200004];Long Longs[200004];intMain () {Long LongN,tot; CIN>>n>>tot; Long Longk=tot-(n1); sum[0]=0;  for(Long LongI=1; i<=n;i++) {cin>>S[i]; if(s[i]>=k) Sum[i]=sum[i-1]+K; ElseSum[i]=sum[i-1]+S[i]; }     Long Longfirst=0;  for(Long LongI=1; i<=n;i++)    {       Long Longans=0; if(s[i]>k) Ans+ = (s[i]-k); Long Longtemp=tot-(sum[n]-(sum[i]-sum[i-1])); if(temp-1<=0) Temp=1; Ans+=min (temp-1, S[i]); if(first++) cout<<" "; cout<<ans; } cout<<Endl;}

D: Starting from 0, if the current value is >=3, but not found in the sequence, then only 3, that is, only the number of handshakes +1 and the number of people shaking hands -3 of the two cases. If it is still not found, the output impossible.

#include <iostream>#include<string>#include<queue>using namespaceStd;queue<int> q[200005];intres[200005];intMain () {intN,num; CIN>>N;  for(intI=1; i<=n;i++) {cin>>num;    Q[num].push (i); }    intCur=0;  for(intI=1; i<=n;i++)    {         while(Q[cur].empty () &&cur>=3) cur-=3;  while(Q[cur].empty ()) {cout<<"Impossible"<<Endl; return 0; }        inttemp=Q[cur].front ();        Q[cur].pop (); Res[i]=temp; Cur++; }    intfirst=0; cout<<"Possible"<<Endl;  for(intI=1; i<=n;i++)    {        if(first++) cout<<" "; cout<<Res[i]; } cout<<Endl; }

E: Waiting for updates

Codeforces #298题解