結果

問題 No.765 ukuku 2
ユーザー rickytheta
提出日時 2018-12-13 00:56:34
言語 C++11(廃止可能性あり)
(gcc 13.3.0)
結果
AC  
実行時間 134 ms / 3,000 ms
コード長 3,653 bytes
コンパイル時間 1,624 ms
コンパイル使用メモリ 163,104 KB
実行使用メモリ 7,112 KB
最終ジャッジ日時 2024-09-25 03:54:13
合計ジャッジ時間 5,168 ms
ジャッジサーバーID
(参考情報)
judge2 / judge5
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 2
other AC * 48
権限があれば一括ダウンロードができます
コンパイルメッセージ
main.cpp: In function ‘int main()’:
main.cpp:28:8: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
   28 |   scanf("%s",s);
      |   ~~~~~^~~~~~~~

ソースコード

diff #
プレゼンテーションモードにする

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define REP(i,n) for(int i=0;i<(int)(n);i++)
#define FOR(i,a,b) for(int i=(int)(a);i<(int)(b);i++)
#define FORR(i,a,b) for(int i=(int)(b)-1;i>=(int)(a);i--)
#define CHMIN(a,b) (a)=min((a),(b))
#define CHMAX(a,b) (a)=max((a),(b))
// const ll B = 10007;
const ll B = 10007;
const ll MOD = 1000000009;
int bpow[252521];
int n;
char s[252521];
char rs[252521];
int hsh[252521], rhsh[252521];
int rhash(int *h, int l, int r){
// s[l,...,r-1]
return (h[r] - (ll)h[l]*bpow[r-l]%MOD + MOD) % MOD;
}
int main(){
scanf("%s",s);
n = strlen(s);
REP(i,n)rs[n-1-i] = s[i];
REP(i,n)hsh[i+1] = ((ll)hsh[i]*B + s[i]) % MOD;
REP(i,n)rhsh[i+1] = ((ll)rhsh[i]*B + rs[i]) % MOD;
bpow[0] = 1;
FOR(i,1,252521)bpow[i] = (ll)bpow[i-1]*B%MOD;
int ans = 1;
// center character is s[i]
REP(i,n){
// no remove
int low = 0, high = min(i, n-1-i)+1;
while(low+1<high){
int len = (low+high)/2;
int lefthash = rhash(hsh, i-len, i);
int righthash = rhash(rhsh, n-1-len-i, n-i-1);
if(lefthash == righthash){
low = len;
}else{
high = len;
}
}
int clen = low;
CHMAX(ans, 2*clen); // remove center character
if(2*clen+1 < n)CHMAX(ans, 2*clen+1); // remove other character
// remove left
if(i-clen > 0){
int l = i-clen-1;
int r = i+clen;
int low = 0, high = min(l, n-1-r)+1;
while(low+1<high){
int len = (low+high)/2;
int lefthash = rhash(hsh, l-len, l);
int righthash = rhash(rhsh, n-1-len-r, n-r-1);
if(lefthash == righthash){
low = len;
}else{
high = len;
}
}
CHMAX(ans, 2*clen+2*low+1);
}
// remove right
if(i+clen < n-1){
int l = i-clen;
int r = i+clen+1;
int low = 0, high = min(l, n-1-r)+1;
while(low+1<high){
int len = (low+high)/2;
int lefthash = rhash(hsh, l-len, l);
int righthash = rhash(rhsh, n-1-len-r, n-r-1);
if(lefthash == righthash){
low = len;
}else{
high = len;
}
}
CHMAX(ans, 2*clen+2*low+1);
}
}
// center character is s[i], s[i+1]
REP(i,n-1)if(s[i]==s[i+1]){
// no remove
int l = i, r = i+1;
int low = 0, high = min(l, n-1-r)+1;
while(low+1<high){
int len = (low+high)/2;
int lefthash = rhash(hsh, l-len, l);
int righthash = rhash(rhsh, n-1-len-r, n-r-1);
if(lefthash == righthash){
low = len;
}else{
high = len;
}
}
int clen = low;
CHMAX(ans, 2*clen+1); // remove center character
if(2*clen+2 < n)CHMAX(ans, 2*clen+2); // remove other character
// remove left
if(i-clen > 0){
int l = i-clen-1;
int r = i+1+clen;
int low = 0, high = min(l, n-1-r)+1;
while(low+1<high){
int len = (low+high)/2;
int lefthash = rhash(hsh, l-len, l);
int righthash = rhash(rhsh, n-1-len-r, n-r-1);
if(lefthash == righthash){
low = len;
}else{
high = len;
}
}
CHMAX(ans, 2*clen+2+2*low);
}
// remove right
if(i+1+clen < n-1){
int l = i-clen;
int r = i+1+clen+1;
int low = 0, high = min(l, n-1-r)+1;
while(low+1<high){
int len = (low+high)/2;
int lefthash = rhash(hsh, l-len, l);
int righthash = rhash(rhsh, n-1-len-r, n-r-1);
if(lefthash == righthash){
low = len;
}else{
high = len;
}
}
CHMAX(ans, 2*clen+2+2*low);
}
}
printf("%d\n",ans);
return 0;
}
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