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20
2016
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Diary of OI

Jan.20th

注:黑粗体为超链接
照题目要求模拟就行了
裸的扩欧
没有学过的童鞋戳右边→_→扩展欧几里得算法详解
 C++ Code 
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#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cctype>
#include<ctime>
#include<cstdlib>
#include<string>
#include<queue>
#include<cmath>
#define Rep(x,a,b) for (int x=a;x<=b;x++)
#define Per(x,a,b) for (int x=a;x>=b;x--)
#define ll long long
using namespace std;
inline int IN(){
    int x=0,ch=getchar(),f=1;
    while (!isdigit(ch)&&(ch!='-')&&(ch!=EOF)) ch=getchar();
    if (ch=='-'){f=-1;ch=getchar();}
    while (isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
    return x*f;
}
int exgcd(int a,int b,int &x,int &y){
    if (b==0){x=1;y=0;return a;}
    int gcd=exgcd(b,a%b,x,y);
    int t=x;
    x=y;y=t-a/b*y;
    return gcd;
}
int A,B,C;
int main(){
    A=IN(),B=IN(),C=-1*IN();
    int x,y;
    int gcd=exgcd(A,B,x,y);
    if (C%gcd!=0return puts("-1"),0;
    ll s=C/gcd;
    printf("%I64d %I64d\n",s*x,s*y);
}
主要就是要在线性时间内求出所有的前缀是不是回文串
这个我们可以用扩展KMP搞
没有学过的童鞋戳右边→_→扩展KMP算法详解
然后我们把串S读进来 再反一下记为T
再把串T的后缀们与S做一下扩展KMP
如果对于串T存在$ex_i+i=len$那么也就是说:
在串S中$[ 0 , len-i+1 ]$这个前缀是回文串
然后剩下的就比较傻了
 C++ Code 
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#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cctype>
#include<ctime>
#include<cstdlib>
#include<string>
#include<queue>
#include<cmath>
#define Rep(x,a,b) for (int x=a;x<=b;x++)
#define Per(x,a,b) for (int x=a;x>=b;x--)
#define ll long long
using namespace std;
inline int IN(){
    int x=0,ch=getchar(),f=1;
    while (!isdigit(ch)&&(ch!='-')&&(ch!=EOF)) ch=getchar();
    if (ch=='-'){f=-1;ch=getchar();}
    while (isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
    return x*f;
}
const int maxlen=5000005;
char s[maxlen],t[maxlen];
int next[maxlen],ex[maxlen],len,lev[maxlen];
void GetNext(){
    next[0]=len;
    int k=0;
    while (k<len-1&&s[k]==s[k+1]) k++;
    next[1]=k;k=1;
    Rep(i,2,len-1){
        int p=k+next[k]-1,L=next[i-k];
        if (i+L<=p) next[i]=L;
        else{
            int j=max(p-i+1,0);
            while (i+j<len&&s[j]==s[i+j]) j++;
            next[i]=j;
            k=i;
        }
    }
}
void GetExtend(){
    int k=0;
    while (k<len&&t[k]==s[k]) k++;
    ex[0]=k;k=0;
    Rep(i,1,len-1){
        int p=k+ex[k]-1,L=next[i-k];
        if (i+L<=p) ex[i]=L;
        else{
            int j=max(p-i+1,0);
            while (i+j<len&&s[j]==t[i+j]) j++;
            ex[i]=j;
            k=i;
        }
    }
}
int main(){
    gets(s);
    len=strlen(s);t[len]=0;
    Rep(i,0,len-1) t[i]=s[len-i-1];
    GetNext();GetExtend();
    int Ans=0;
    Rep(i,0,len-1){
        if (ex[len-i-1]+len-i-1==len) lev[i]=lev[(i-1)/2]+1,Ans+=lev[i];
        else lev[i]=0;
    }
    printf("%d\n",Ans);
}

 

Category: Diary | Tags: codeforces | Read Count: 451
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