이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include<bits/stdc++.h>
using namespace std;
const int nmax=2e6+42,base=31,mod=1e9+9;
string inp;
string outp;
int n,sz;
int st[nmax],pref[nmax];
set<int> seen;
int ask(int l,int r)
{
if(l>r)return 0;
long long help=pref[r];
long long sub=0;
if(l)sub=pref[l-1];
sub=sub*st[r-l+1]%mod;
//cout<<"ask "<<l<<" "<<r<<" -> "<<((help-sub)%mod+mod)%mod<<endl;
return ((help-sub)%mod+mod)%mod;
}
bool been_0=0,been_n=0;
void pick(int l,int r)
{
if(l==0&&been_0)return;
if(l==n+1&&been_n)return;
string help="";
for(int i=l;i<=r;i++)
help.push_back(inp[i]);
if(outp.size())
{
if(help!=outp)
{
cout<<"NOT UNIQUE"<<endl;
exit(0);
}
}
outp=help;
if(l==0)been_0=1;
if(l==n+1)been_n=1;
}
int main()
{
scanf("%i",&sz);
if(sz%2==0)
{
cout<<"NOT POSSIBLE"<<endl;
return 0;
}
for(int i=0;i<sz;i++)
{
char c=getchar();
while('A'>c||c>'Z')c=getchar();
inp.push_back(c);
}
n=(sz-1)/2;
pref[0]=inp[0]-'A';
for(int i=1;i<sz;i++)
pref[i]=(1LL*pref[i-1]*base+inp[i]-'A')%mod;
st[0]=1;
for(int i=1;i<sz;i++)st[i]=1LL*st[i-1]*base%mod;
int mem=ask(n+1,sz-1);
//in the first half
for(int i=0;i<=n;i++)
{
//cut i-th
long long part1=1LL*pref[i-1]*st[n-i];
long long part2=ask(i+1,n);
part1=(part1+part2)%mod;
//cout<<part1<<" "<<ask(n+1,sz-1)<<endl;
if(part1==mem)
pick(n+1,sz-1);
}
mem=ask(0,n-1);
//in the second half
for(int i=n;i<sz;i++)
{
//cut i-th
long long part1=1LL*ask(n,i-1)*st[sz-1-i];
long long part2=ask(i+1,sz-1);
part1=(part1+part2)%mod;
//cout<<part1<<" "<<ask(0,n-1)<<endl;
if(part1==mem)
pick(0,n-1);
}
if(outp.size())
{
for(auto k:outp)
printf("%c",k);
printf("\n");
}
else cout<<"NOT POSSIBLE"<<endl;
return 0;
}
컴파일 시 표준 에러 (stderr) 메시지
friends.cpp: In function 'int main()':
friends.cpp:59:10: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
scanf("%i",&sz);
~~~~~^~~~~~~~~~
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