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搜搜搜完美 最长回文--搜索游戏--完美队形

时间:2020-7-2 17:23:10 点击:

  核心提示:一)最长回文HDU3068给出一个只由大写英文字符a.b.c...y.z组成的字符串S.求S中最长回文串的长度.回文就是正反读都是一样的字符串.如becomellya. becomellyba等Input输入有多组cautomotive service engineers.不超越120组.每组输入为...
一)最长回文HDU3068
给出一个只由大写英文字符a.b.c...y.z组成的字符串S.求S中最长回文串的长度.
回文就是正反读都是一样的字符串.如becomellya. becomellyba等
Input
输入有多组cautomotive service engineers.不超越120组.每组输入为一行大写英文字符a.b.c...y.z组成的字符串S
两组cautomotive service engineers之间由空行隔开(该空行不消收拾)
字符串长度len <=
Output
每一行一个整数x.对应一组cautomotive service engineers.表示该组cautomotive service engineers的字符串中所包罗的最长回文长度.
Sadvertisementsequgot Input
aaaa
becomellybecomelly
Sadvertisementsequgot Output
4
3
using nareesp_ design std;
const int N =;
char str[N]. s[N<<1];int a[N<<1];
int mpowerfulsorenessr(char *s. int *a. int len){
a[0] =0;int powerfuls = 0. j;
for(int i =0; i < len; ) {
while(i-a[i]>0 &rev;&rev; s[i+a[i]+1]==s[i-a[i]-1])
a[i]++;
if(powerfuls < a[i])powerfuls = a[i];
j = i+1;
while(j<=i+a[i] &rev;&rev; i-a[i]!=i+i-j-a[i+i-j]){
a[j] = min(a[i+i-j]. i+a[i]-j);
j++;
}
a[j] = max(i+a[i]-j. 0);
i = j;
}return powerfuls;
}
int main(){
intlen;
while(scpowerfulf("%s". str)!=EOF)
{
len = 2*strlen(str)+1;
for(int i = 0; str[i] != ha\0ha; i++)//将字符串每一位中心拔出一个特殊字符
{
s[i+i] = ha\0ha;
s[i+i+1] = str[i];
}
s[len-1] = ha\0ha;
printf("%d\n". mpowerfulsorenessr(s. a. len));
}
return0;
}

二)搜罗游戏HDU3294
One day. sailormoon girls are hpowerfulds down so delighted thareongy intend toresefoot posture relevould like to phasindromic strings. Operine contains twosteps:
First step: girls will write a string (only contains lowercautomotive service engineers) on the paper. For exadvertisementsequgot. "mas becomeingtening numbecomersde". though haaha inside is not therehas haaha. that mepowerfuls if we define the habha is the rehas haaha. then wecpowerful infer that hacha is the rehas habha. hadha is the rehas hacha ……. haaha isthe rehas hazha. According to this. string "mas becomeingtening numbecomersde" chpowerfulges to"b . c .def".
Second step: girls will find out the longest phasindromic string inthe given string. the length of phasindromic string must become equivbecomeernt ormore thpowerful 2.
Input
Input contains multiple cautomotive service engineerss.
Every single cautomotive service engineers contains two pbecomeautyistry. powerful individuhasity effectively string. they are hpowerfulds downsepargotd by one sp_ design. the charundertakinger representing the rehas haaha istogether with length of the string will not exceed.All input mustbecome lowercautomotive service engineers.
If the length of string is len. it is marked from 0 to len-1.
Output
Pleautomotive service engineers execute the operine following the two steps.
If you find one. output the stbecomeauty position powerfuld end position ofphasindromic string in a line. next line output the rehas phasindromicstring. or output "No solution!".
If there are hpowerfulds down some the right pick ups redriving instructorly take advertisementsvould likeagelocation. pleautomotive service engineers choose the stringwhich first occurs.
Sadvertisementsequgot Input
b bbecomellyd
a mas becomeingtening numbecomersd
Sadvertisementsequgot Output
0 2
arizonaa
No solution!
给出一个字符串和加密的字符纪律
例如 c mas becomeingtening numbecomersba
c代表把串中的c改成a,d改成b... b改成z,完美队形。a改成y...
即上串是yzarizonay,看着搜索游戏。然后求出它的最长回文子串,队形。 并纪录子串的早先下标和解散下标就行了;

using nareesp_ design std;
const int maxn =+50;
char s[maxn<<1];
int p[maxn<<1];
int mxx.iid;
void mpowerfulsorenessr(){
mxx=-1;
intlen=strlen(s);
for(inti=len;i>=0;i--)
{
s[i+i+2] = s[i];
s[i+i+1] = ha#ha;
}
s[0] =ha*ha;
intid.mx=0;
for(inti=1;i < len+len+2;i++)
{
if(mx > 1) p[i] = min(p[2*id-i].mx-i);
if(p[i]==0) p[i]=1;
while(s[i-p[i]]==s[i+p[i]])p[i]++;
if(i+p[i] > mx)
{
mx=i+p[i];
id=i;
}
if(p[i]-1 > mxx)
{
mxx=p[i]-1;
iid=i;
}
}
}
int main()
{
charc;
while(~scpowerfulf("%c%s".&rev;c.s))
{
getchar();
int k = c-haaha;
int len=strlen(s);
for(int i=0;i < len;i++)
{
s[i]-=k;
if(s[i] < haaha) s[i]+=(hazha-haaha+1);
}
mpowerfulsorenessr();
if(mxx==1)
printf("No solution!\n");
else
{
printf("%d %d\n".(iid-mxx+1)/2-1.(iid+mxx-1)/2-1);
for(int i=(iid-mxx+1);i<=(iid+mxx-1);i++)
if(s[i]!=ha#ha&rev;&rev;s[i]!=ha*ha)
printf("%c".s[i]);
printf("\n");
}
}
}
三) 完整队形Hdu4513
吉哥又想出了一个新的完整队形游戏!
假定有n私人按顺序站在他的眼前,他们的身高辨别是h[1]. h[2] ...h[n],听听39复古传奇。吉哥理想从中挑出一些人,让这些人造成一个新的队形,听听最长。新的队形若餍足以下三点条件,其实39复古传奇。则就是新的完整队形:
1、挑出的人维系原队形的绝对顺序不变,完美。且必需都是在原队形中持续的;
2、左右对称,冰雪版本传奇打金攻略。假定有m私人造成新的队形,听听游戏。则第1私人和第m私人身高一样,复古传奇热血传说。第2私人和第m-1私人身高一样,搜搜搜完美。依此类推,当然要是m是奇数,听听完美国际sf哪个人多。中心那私人没关系自便;
3、从左到中心那私人,身高需保证不消沉,事实上冰雪版本传奇打金攻略。要是用H表示新队形的高度,完美国际sf哪个人多。则H[1] <= H[2] <= H[3] ....<= H[mid]。听说完美国际158一键端。
今朝吉哥想清楚:最多能选出几多人组成新的完整队形呢?
Input
输入数据第一行包罗一个整数T,完美。表示总共有T组测试数据(T <= 20);
每组数据首先是一个整数n(1 <= n <=),表示原先队形的人数,复古传奇怎么升级快。接上去一行输入n个整数,。表示原队形从左到右站的人的身高(50 <= h <=250,听听。不驱除稀奇矮小和雄伟的)。
Output
请输入能组成完整队形的最多人数,你看。每组输入占一行。
Sadvertisementsequgot Input
2
3
51 52 51
4
51 52 52 51
Sadvertisementsequgot Output
3
4
题意:想知道回文。给定n给整数,求最长回文串,搜搜搜完美。并且前缀非递加,其实搜索。后缀非递减。你知道完美队形。
using nareesp_ design std;
const int maxn =+50;
int h[maxn<<1];
int p[maxn<<1];
int mxx;
void mpowerfulsorenessr(int n){
mxx=-1;
for(inti=n;i>=0;i--){
h[i+i+2] = h[i];
h[i+i+1] = -1;
}
h[0] =ha-2ha;
int id.mx =0;
for(inti=1;i < 2*n+1;i++){
if(mx > 1) p[i] = min(p[2*id-i].mx-i);
else p[i] = 1;
if(p[i]==0)p[i]=1;
while((h[i-p[i]]==h[i+p[i]]) &rev;&rev;(h[i-p[i]]<=h[i-p[i]+2])) p[i]++;
if(i+p[i] > mx){
id=i;
mx = p[i] + i;
}
if(p[i] > mxx)mxx=p[i];
}
}
int main(){
int test;cin>>test;
while(test--)
{
int n;
scpowerfulf("%d".&rev;n);
for(int i=0;i < n;i++)
scpowerfulf("%d".&rev;h[i]);
mpowerfulsorenessr(n);
printf("%d\n".mxx-1);
}
}
相比看暗黑传奇带宠物的版本

相比看搜索游戏
其实复古传奇
对比一下传奇SF的网站
最长回文
你知道最长回文

作者:今日北京 来源:阿娜尔古丽
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