KMP & 扩展KMP & Manacher

A Number Sequence

求b序列在a序列出现的次数，KMP模板题：

#include <iostream>
#include <string.h>
#include <stdio.h>
char str[10010],str1[1000010];
int nex[10010];
void init(){
    int j = nex[0] = -1, i = 0;
    int len = strlen(str);
    while(i < len){
        if(j == -1 || str[i] == str[j]){
            nex[++i] = ++j;
        }else j = nex[j];
    }
}
int KMP(){
    int i  = 0, j = 0, sum = 0;
    int len = strlen(str), len1 = strlen(str1);
    while(j < len1){
        if(i == -1 || str[i] == str1[j]){
            i++;j++;
        }else i = nex[i];
        if(i == len) sum++;
    }
    return sum;
}
using namespace std;
int main(){
    int n;
    scanf("%d",&n);
    while(n--){
        scanf("%s%s",str,str1);
        init();
        /*for(int i = 0; i <= strlen(str); i ++){
            printf("%d ",nex[i]);
        }
        printf("
");*/
        printf("%d
",KMP());
        memset(str,0,sizeof(str));
        memset(str1,0,sizeof(str1));
        memset(nex,0,sizeof(nex));
    }
    return 0;
}

B - Oulipo

求第一个字符串在第二个字符串中出现的次数，KMP模板题：

#include <iostream>
#include <string.h>
#include <stdio.h>
char str[10010],str1[1000010];
int nex[10010];
void init(){
    int j = nex[0] = -1, i = 0;
    int len = strlen(str);
    while(i < len){
        if(j == -1 || str[i] == str[j]){
            nex[++i] = ++j;
        }else j = nex[j];
    }
}
int KMP(){
    int i  = 0, j = 0, sum = 0;
    int len = strlen(str), len1 = strlen(str1);
    while(j < len1){
        if(i == -1 || str[i] == str1[j]){
            i++;j++;
        }else i = nex[i];
        if(i == len) sum++;
    }
    return sum;
}
using namespace std;
int main(){
    int n;
    scanf("%d",&n);
    while(n--){
        scanf("%s%s",str,str1);
        init();
        /*for(int i = 0; i <= strlen(str); i ++){
            printf("%d ",nex[i]);
        }
        printf("
");*/
        printf("%d
",KMP());
        memset(str,0,sizeof(str));
        memset(str1,0,sizeof(str1));
        memset(nex,0,sizeof(nex));
    }
    return 0;
}

C - 剪花布条

求第二个字符串在第一个字符串中出现的次数，不能重复，KMP中每次找到让其下边重置为0就行了。

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
char str[1010],str1[1010];
int nex[1010];
void init(){
    int j = nex[0] = -1, i = 0;
    int len1 = strlen(str1);
    while(i < len1){
        if(j == -1 || str1[i] == str1[j]){
            nex[++i] = ++j;
        }else j = nex[j];
    }
}
int KMP(){
    int i = 0,j = 0, ans = 0;
    int len = strlen(str), len1 = strlen(str1);
    while(i < len){
        if(j == -1 || str[i] == str1[j]){
            i++;j++;
            if(j == len1){
                ans++;j=0;
            }
        }else j = nex[j];
    }
    return ans;
}

int main(){
    while(scanf("%s",str)!=EOF){
        if(!strcmp(str,"#"))break;
        scanf("%s",str1);
        init();
        printf("%d
",KMP());
        memset(str,0,sizeof(str));
        memset(str1,0,sizeof(str1));
    }
    return 0;
}

D - Cyclic Nacklace

这题就是求循环节。

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
char str[100010];
int nex[100010];
void init(){
    int j = nex[0] = -1, i = 0;
    int len = strlen(str);
    while(i < len){
        if(j == -1 || str[i] == str[j]){
            nex[++i] = ++j;
        }else j = nex[j];
    }
}
int KMP(){
    int len = strlen(str);
    int ans = len -nex[len];
    if(!nex[len])return len;
    else if(len%ans) return ans-len%ans;
    else return 0;
}
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%s",str);
        init();
        printf("%d
",KMP());
    }
    return 0;
}

E - Period

输出一遍next数组就看出规则了，直接过了。

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
const int MAX = 1e6+10;
int n, nex[MAX];
char str[MAX];
void init(){
    int j = nex[0] = -1, i = 0;
    int len = strlen(str);
    while(i < len){
        if(j == -1 || str[i] == str[j]){
            nex[++i] = ++j;
        }else j = nex[j];
    }
}
void KMP(){
    int len = strlen(str);
    for(int i = 2; i <= len; i ++){
        int flag = 0, ans = i-nex[i];
        if(i%ans || ans == i)continue;
        printf("%d %d
",i,i/(i-nex[i]));
    }
    printf("
");
}
int main(){
    int k = 1;
    while(scanf("%d",&n)!=EOF){
        if(!n)break;
        scanf("%s",str);
        printf("Test case #%d
",k++);
        init();
        /*for(int i = 0; i <= strlen(str); i ++){
            printf("%d ",nex[i]);
        }
        printf("
");*/
        KMP();
    }
    return 0;
}

F - Power Strings

关于循环节的问题。

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
const int MAX = 1e6+10;
char str[MAX];
int nex[MAX];
void init(){
    int j = nex[0] = -1, i = 0;
    int len = strlen(str);
    while(i < len){
        if(j == -1 || str[i] == str[j]){
            nex[++i] = ++j;
        }else j = nex[j];
    }
}
int main(){
    while(scanf("%s",str)!=EOF){
        if(str[0] == '.')break;
        init();
        int len = strlen(str);
        int ans = len - nex[len];
        if(len%ans==0)
            printf("%d
",len/ans);
        else printf("1
");
        memset(str,0,sizeof(str));
        memset(nex,0,sizeof(nex));
    }
    return 0;
}

G - Seek the Name, Seek the Fame

求循环的位置，用数组保存下输出就行了。

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
char str[400010];
int nex[400010];
int a[400010];
void init(){
    int j = nex[0] = -1, i = 0;
    int len = strlen(str);
    while(i < len){
        if(j == -1 || str[i] == str[j]) nex[++i] = ++j;
        else j = nex[j];
    }
}
int main(){
    while(scanf("%s",str)!=EOF){
        init();
        int k = 0,len = strlen(str);
        int i = len;
        while(nex[i]){
            a[k++] = nex[i];
            i = nex[i];
        }
        for(int i = k-1; i >= 0; i --){
            printf("%d ",a[i]);
        }
        printf("%d
",len);
    }
    return 0;
}

H - Blue Jeans

直接暴力就可以做了。

#include<cstdio>
#include<string.h>
#include<iostream>
using namespace std;

int main(){
    int t,n,i,j;
    string str[11];
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        for(i=0;i<n;i++)
            cin>>str[i];
        string ans;
        for(i=3;i<=60;i++){
            for(j=0;j<=60-3;j++){
                bool flag = true;
                string s=str[0].substr(j,i);
                for(int k = 1; k < n; k ++)
                    if(str[k].find(s) == string::npos){
                        flag = false;
                        break;
                    }
                if(flag && s.size() > ans.size())
                    ans = s;
                else if(flag && s.size() == ans.size() && s < ans)
                    ans = s;
            }
        }
        if(!ans.empty())
            printf("%s
",ans.c_str());
        else 
            printf("no significant commonalities
");
    }
    return 0;
}

I - Simpsons’ Hidden Talents

扩展KMP模板题。

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
char str1[50010],str2[50010];
int nex[50010],extend[50010];

void init(){
    int len = strlen(str1);
    nex[0] = len;
    int a, p;
    for(int i = 1,j=-1; i < len; i++,j--){
        if(j < 0 || i+nex[i-a] >= p){
            if(j < 0)
                p = i,j = 0;
            while(p < len && str1[p] == str1[j])
                p++,j++;
            nex[i] = j;
            a = i;
        }else nex[i] = nex[i-a];
    }
}
void getExtend(){
    int a,p;
    int len1 = strlen(str1), len2 = strlen(str2);
    for(int i = 0,j=-1; i < len2; i++,j--){
        if(j < 0 || i+nex[i-a]>=p){
            if(j<0)
                p=i,j=0;
            while(p<len2 && j<len1 && str2[p] == str1[j])
                p++,j++;
            extend[i] = j;
            a = i;
        }else extend[i] = extend[i-a];
    }
}
int main(){
    while(scanf("%s%s",str1,str2)!=EOF){
        init();
        getExtend();
        int i, len2 = strlen(str2);
        for(i = 0; i < len2; i ++){
            if(i+extend[i] == len2){
                printf("%s %d
",str2+i,extend[i]);
                break;
            }
        }
        if(i == len2)printf("0
");
    }
    return 0;
}

J - Count the string   （***）

主要是要理解以i结尾的子串中含前缀的数量加上前j个字符这一前缀。

#include <iostream>
#include <stdio.h>
#include <string.h>
const int MAX = 200010,Mod = 10007;
char str[MAX];
int nex[MAX],sum[MAX],ans;
void init(){
    int j = nex[0] = -1, i = 0;
    int len = strlen(str);
    while(i < len){
        if(j == -1 || str[i] == str[j]){
            if(j!=-1){
                sum[i] = sum[j]+1;
                ans+=sum[i];
            }
            nex[++i] = ++j;
        }
        else j = nex[j];
    }
}
int main(){
    int t,n;
    scanf("%d",&t);
    while(t--){
        scanf("%d%s",&n,str);
        ans = 0;
        init();
        printf("%d
",(ans+n)%Mod);
        memset(str,0,sizeof(str));
        memset(sum,0,sizeof(sum));
    }
    return 0;
}

 

K - Clairewd’s message

这题理解好久才懂题意，就是第一行给你一个有26字符的字符串，分别对于a,b,c,,z,。第二行字符串中是密文加明文，密文要大于等于明文，明文可能也没有，要你输出密文加完整的明文，由于密文都有，求出明文就好办了。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <map>
using namespace std;
const int MAX = 100010;
char str[MAX],s[30],str1[MAX];
int nex[MAX],extend[MAX];
map<char,char> mp;
void init(){
    int len = strlen(str1),a,p;
    nex[0] = len;
    for(int i = 1, j = -1; i < len; i ++, j--){
        if(j < 0 || i+nex[i-a] >= p){
            if(j < 0)
                p=i,j=0;
            while(p < len && str1[p]==str1[j])
                p++,j++;
            nex[i] = j;
            a = i;
        }else nex[i] = nex[i-a];
    }
}
void KMP(){
    int a,p;
    int len = strlen(str), len1 = strlen(str1);
    for(int i = 1, j = -1; i < len; i++,j--){
        if(j<0 || i+nex[i-a] >= p){
            if(j < 0)
                p=i,j=0;
            while(p < len && str[p]==str1[j])
                p++,j++;
            extend[i] = j;
            a = i;
        }else extend[i] = nex[i-a];
    }
}
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%s%s",s,str);
        for(int i = 0;i < 26; i ++)
            mp[s[i]] = 'a'+i;
        for(int i = 0; str[i]; i ++)
            str1[i] = mp[str[i]];
        printf("%s",str,str1);
        init();
        KMP();
        int k,len=strlen(str);
        //printf("
------%s---------
",str1);
        for(k = (len+1)/2; k < len; k++){
            if(k+extend[k] == len)
                break;
        }
        //printf("
+++++++++%d++++++
",k);
        for(int i =len-k ;i < k; i++){
            putchar(str1[i]);
        }
        printf("
");
    }
    return 0;
}

L - Substrings

求最大的公共子串，暴力可以解决。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

int main(){
    int t,n,i,j,k;
    scanf("%d",&t);
    string str[110];
    while(t--){
        scanf("%d",&n);
        for(i=0;i<n;i++)
            cin>>str[i];
        int ans = 0;
        for(i = str[0].size();i > 0; i--){
            for(j = 0; j <= str[0].size()-i; j++){
                string s = str[0].substr(j,i);
                string ss = s;
                reverse(s.begin(),s.end());
                for(k = 1; k < n; k ++)
                    if(str[k].find(s,0) == -1 && str[k].find(ss,0) == -1){
                        break;
                    }
                if(k == n && s.size() > ans)
                    ans = s.size();
            }
        }
        printf("%d
",ans);
    }
    return 0;
}

M - Corporate Identity

和L题差不多，暴力。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

int main(){
    int n;
    string str[4002];
    while(scanf("%d",&n)&&n){
        int cnt = 220;
        string s, ans;
        for(int i = 0; i < n; i++){
            cin>>str[i];
            if(str[i].size() < cnt){
                cnt = str[i].size();
                s = str[i];
            }
        }
        for(int i = cnt; i > 0; i--){
            for(int j = 0; j <= cnt-i; j++){
                string ss = s.substr(j,i);
                int k;
                for(k = 0; k < n; k ++){
                    if(str[k].find(ss,0) == -1)
                        break;
                }
                if(k == n && ss.size() > ans.size())
                    ans = ss;
                else if(k==n && ss.size() == ans.size() && ans > ss)
                    ans = ss;
            }
        }
        if(!ans.empty())
            printf("%s
",ans.c_str());
        else printf("IDENTITY LOST
");
    }
    return 0;
}

N - String Problem

求最小字典和最大字典，并求出他们的数量，用最小表示法可以快速求最先字典，同理也可以求最大字典。次数就是循环节了。

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
const int MAX = 1000010;
char str[MAX];
int nex[MAX];
void init(){
    int j = nex[0] = -1, i = 0;
    int len = strlen(str);
    while(i < len){
        if(j == -1 || str[i] == str[j]) nex[++i] = ++j;
        else j = nex[j];
    }
}
int minAndMax(bool flag){
    int i = 0, j = 1, k = 0;
    int len = strlen(str);
    while(i<len && j<len && k<len){
        int t = str[(j+k)%len]-str[(i+k)%len];
        if(t == 0) k++;
        else {
            if(flag){
                if(t>0)j+=k+1;
                else i+=k+1;
            }else{
                if(t>0)i+=k+1;
                else j+=k+1;
            }
            if(i==j)j++;
            k=0;
        }
    }
    return min(j,i);
}
int main(){
    while(scanf("%s",str)!=EOF){
        init();
        int MIN = minAndMax(true);
        int MAX = minAndMax(false);
        int len = strlen(str);
        int cnt = len-nex[len];
        int ans = len%cnt?1:len/cnt;
        printf("%d %d %d %d
",MIN+1,ans,MAX+1,ans);
        memset(str,0,sizeof(str));
    }
    return 0;
}

O - How many

最小表示法加set可以解决。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <set>
using namespace std;
char str[110];
set<string> ss;
int Min_index(char *ss){
    int i = 0,j = 1, k = 0;
    int len = strlen(ss);
    while(i<len && j<len && k<len){
        int t = ss[(j+k)%len]-ss[(i+k)%len];
        if(t == 0)k++;
        else{
            if(t>0) j+=k+1;
            else i+=k+1;
            k=0;
        }
        if(i==j)j++;
    }
    return min(i,j);
}
int main(){
    int n;
    while(scanf("%d",&n)!=EOF){
        string s,ans;
        for(int i = 0; i < n; i ++){
            scanf("%s",str);
            int k = Min_index(str);
            ans = str;
            s = ans.substr(k,ans.length()-k)+ans.substr(0,k);
            ss.insert(s);
        }
        printf("%d
",ss.size());
        ss.clear();
    }
    return 0;
}

P - Period II

next数组的运用。

#include <iostream>
#include <stdio.h>
#include <string.h>
const int MAX = 1000010;
char str[MAX];
int nex[MAX],a[MAX];
void init(){
    int len = strlen(str),a,p;
    nex[0] = len;
    for(int i = 1,j = -1; i < len; i ++,j--){
        if(j<0 || i+nex[i-a] >= p){
            if(j < 0)
                p=i,j=0;
            while(p<len && str[p]==str[j])
                p++,j++;
            nex[i]=j;
            a=i;
        }else nex[i] = nex[i-a];
    }
}
int main(){
    int n;
    while(scanf("%d",&n)!=EOF){
        for(int i = 1; i <= n; i ++){
            scanf("%s",str);
            init();
            int k = 0,len = strlen(str);
            for(int j = 1; j < len; j++){
                if(j+nex[j] == len)
                    a[k++] = j;
            }
            printf("Case #%d: %d
",i,k+1);
            for(int j = 0; j < k; j ++)
                printf("%d ",a[j]);
            printf("%d
",len);
        }
    }
    return 0;
}

R - Best Reward

将字符串分成两半，不是回文的价值为0，是回文的每个字符代表一个价值，求最大价值，Manacher 的运用（***）。

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
const int MAX = 500010;
int t,v[27],p[MAX*2],sum[MAX],per[MAX],pos[MAX];
char str[MAX*2];

void init(){
    int id = 0,len = strlen(str);
    for(int i = len; i >= 0; i--){
        str[i+i+2] = str[i];
        str[i+i+1] = '#';
    }
    str[0] = '*';
    for(int i = 2; i <2*len+1; i++){
        if(id+p[id] > i) p[i] = min(p[2*id-i],p[id]+id-i);
        else p[i] = 1;
        while(str[i-p[i]] == str[i+p[i]])
            ++p[i];
        if(id+p[id] < p[i]+i) id = i;
        if(i-p[i] == 0)per[p[i]-1] = t+1;
        if(i+p[i] == len*2+2)pos[p[i]-1] = t+1;
    }
}
int main(){
    scanf("%d",&t);
    while(t--){
        for(int i = 0; i < 26; i ++) scanf("%d",&v[i]);
        scanf("%s",str);
        int len = strlen(str);
        for(int i = 0; i < len; i ++){
            sum[i+1] += sum[i]+v[str[i]-'a'];
        }
        init();
        int ans = -1;
        for(int i = 1; i < len; i ++){
            int tmp = 0;
            if(per[i] == t+1)tmp+=sum[i];
            if(pos[len-i] == t+1) tmp+=(sum[len]-sum[i]);
            if(tmp > ans) ans=tmp;
        }
        printf("%d
",ans);
        memset(per,0,sizeof(per));
        memset(pos,0,sizeof(pos));
        memset(sum,0,sizeof(sum));
    }
    return 0;
}

T - Palindrome

Manacher的模板题。

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
const int MAX = 1000010;
char str[MAX<<1];
int p[MAX<<1];
int solve(){
    int ans = 0, id = 0, len = strlen(str);
    for(int i = len; i >= 0; i --){
        str[i+i+2] = str[i];
        str[i+i+1] = '#';
    }
    str[0] = '*';
    for(int i = 2; i<2*len+1; i ++){
        if(p[id] + id > i) p[i] = min(p[2*id-i],p[id]+id-i);
        else p[i] = 1;
        while(str[i-p[i]] == str[i+p[i]])++p[i];
        if(p[id] + id < p[i]+i) id = i;
        if(ans < p[i]) ans = p[i];
    }
    return ans-1;
}
int main(){
    int k = 1;
    while(scanf("%s",str)!=EOF){
        if(str[0] == 'E')break;
        printf("Case %d: %d
",k++,solve());
        memset(str,0,sizeof(str));
        memset(p,0,sizeof(p));
        getchar();
    }
    return 0;
}

U - 吉哥系列故事――完美队形II

Manacher的运用，加了一个条件，从左到中间那个人，身高需保证不下降。

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
const int MAX = 100010;
int a[MAX*2],p[MAX*2],n;

int solve(){
    int ans = 0,id = 0;
    for(int i = n; i >= 0; i --){
        a[i+i+2] = a[i];
        a[i+i+1] = -1;
    }
    a[0] = -2;
    for(int i = 2; i < 2*n+1; i ++){
        if(p[id]+id>i)p[i] = min(p[2*id-i],p[id]+id-i);
        else p[i] = 1;
        while(a[i-p[i]] == a[i+p[i]] && a[i-p[i]] <= a[i-p[i]+2])
            ++p[i];
        if(p[id]+id < p[i]+i)id = i;
        if(ans < p[i]) ans = p[i];
    }
    return ans;
}
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        for(int i = 0; i < n; i ++) scanf("%d",&a[i]);
        printf("%d
",solve()-1);
    }
    return 0;
}

V - Girls' research

求最长回文子串，并输出初末位置和字符串

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
const int MAX = 200010;
char c,str[MAX<<1],gg;
int p[MAX<<1];

void solve(){
    int ans = 0, cnt = 0,id = 0, len = strlen(str);
    for(int i = len; i >= 0; i --){
        str[i+i+2] = str[i];
        str[i+i+1] = '#';
    }
    str[0] = '*';
    for(int i = 2; i<2*len+1; i ++){
        if(p[id] + id > i) p[i] = min(p[2*id-i],p[id]+id-i);
        else p[i] = 1;
        while(str[i-p[i]] == str[i+p[i]])++p[i];
        if(p[id] + id < p[i]+i) id = i;
        if(ans < p[i]) ans = p[i],cnt=i;
    }
    //printf("%d
",p[cnt]);
    if(ans < 3){
        printf("No solution!
");
        return;
    }
    if(str[cnt] == '#'){
         printf("%d %d
",(cnt-p[cnt])/2,(cnt+p[cnt]-4)/2);
         for(int i = (cnt-p[cnt]+2); i <= (cnt+p[cnt]-2); i+=2){
             printf("%c",str[i]);
         }
         printf("
");
    }else{
        printf("%d %d
",(cnt-p[cnt])/2,(cnt+p[cnt]-4)/2);
        for(int i = (cnt-p[cnt]+2); i <= (cnt+p[cnt]-2); i+=2){
            printf("%c",str[i]);
        }
        printf("
");
    }
}

int main(){
    while(scanf("%c %s%c",&c,str,&gg)!=EOF){
        for(int i = 0; str[i]; i ++){
            str[i] = str[i]-c+'a';
            if(str[i] < 'a') str[i] +=26;
        }
        //cout << str << endl;
        solve();
        memset(str,0,sizeof(str));
    }
    return 0;
}

W - 最长回文

Manacher模板题：

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
const int MAX = 110010;
char str[MAX<<1];
int p[MAX<<1];
int solve(){
    int ans = 0, id = 0, len = strlen(str);
    for(int i = len; i >= 0; i --){
        str[i+i+2] = str[i];
        str[i+i+1] = '#';
    }
    str[0] = '*';
    for(int i = 2; i<2*len+1; i ++){
        if(p[id] + id > i) p[i] = min(p[2*id-i],p[id]+id-i);
        else p[i] = 1;
        while(str[i-p[i]] == str[i+p[i]])++p[i];
        if(p[id] + id < p[i]+i) id = i;
        if(ans < p[i]) ans = p[i];
    }
    return ans-1;
}
int main(){
    while(scanf("%s",str)!=EOF){
        printf("%d
",solve());
        memset(str,0,sizeof(str));
        memset(p,0,sizeof(p));
        getchar();
    }
    return 0;
}

X - Wow! Such Doge!

输入输出的处理。

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
int main(){
    char c;
    int ans = 0,d = 0, o = 0, g = 0, e = 0;
    while((c=getchar())!=EOF){
        if(c == ' ' || c == '
'){
            d=o=g=e=0;
            continue;
        }
        if(isalpha(c))
            c = tolower(c);
        
        if(c == 'd'){
            d=1;o=g=e=0;
        }else if(c == 'o' && o==0){
            if(d==1){
                o=1;g=e=0;
            }
        }else if(c == 'g' && g==0){
            if(d==1&& o==1){
                g=1;e=0;
            }
        }else if(c == 'e' && e==0){
            if(d==1&&o==1&&g==1){
                ans++;
                d=o=g=e=0;
            }
        }else {
            d=o=g=e=0;
        }
    }
    printf("%d
",ans);
    return 0;
}

Y - Theme Section

求左边中间和右边最长的公共部分的长度

#include <iostream>
#include <stdio.h>
#include <string.h>
const int MAX = 1e6+10;
char str[MAX];
int nex[MAX],a[MAX/3];
void init(){
    int j = nex[0] = -1, i = 0;
    int len = strlen(str);
    while(i < len){
        if(j == -1 || str[i] == str[j]){
            nex[++i] = ++j;
        }else j = nex[j];
    }
}
int KMP(){
    int len = strlen(str);
    int j = nex[len],i = 0;
    while(j != -1){
        if(j<=MAX/3)
            a[i++] = j;
        j = nex[j];
    }
    if(i == 0 || a[0] == 0) return 0;
    for(int k = 0; k < i; k ++){
        for(int l = a[k]+1; l <= len-a[k]; ++l){
            if(nex[l] == a[k])
                return a[k];
        }
    }
}
int main(){
    int n;
    scanf("%d",&n);
    while(n--){
        scanf("%s",str);
        init();
        printf("%d
",KMP());
        memset(str,0,sizeof(str));
    }
    return 0;
}