SRM 498 DIV1 Easy - FoxSequence

SRM 498 DIV1 Easy - FoxSequence

16:23

問題


http://community.topcoder.com/stat?c=problem_statement&pm=11219&rd=14427

解き方


システムテストで落ちないような実装の正確性が求められる問題。
共通処理は関数化し、各区間について点の不等号がどうであるか
正しく実装する。

コード


class FoxSequence {

public:

bool ispossible(vector<int> seq,int s,int e,int d){
for(int i=s;i<=e;i++)if(seq[s]+(i-s)*d!=seq[i])return false;
return true;
}

string isValid(vector<int> seq) {
int n=seq.size();

int at=0,next=0;
while(next+1<n && seq[next]<seq[next+1])next++;
if(at==next || next>=n)return "NO";
if(!ispossible(seq,at,next,seq[at+1]-seq[at]))return "NO";

at=next;
while(next+1<n && seq[next]>seq[next+1])next++;
if(at==next || next>=n)return "NO";
if(!ispossible(seq,at,next,seq[at+1]-seq[at]))return "NO";

while(next+1<n && seq[next]==seq[next+1])next++;
if(next>=n)return "NO";
at=next;

while(next+1<n && seq[next]<seq[next+1])next++;
if(at==next || next>=n)return "NO";
if(!ispossible(seq,at,next,seq[at+1]-seq[at]))return "NO";

at=next;
while(next+1<n && seq[next]>seq[next+1])next++;
if(at==next || next!=n-1)return "NO";
if(!ispossible(seq,at,next,seq[at+1]-seq[at]))return "NO";

return "YES";
}

};

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