NOIP 模拟 7 考试总结

Posted 2021-06-12 楠枫nan

T1

超级大水题，用 \\(kmp\\) 和 \\(hash\\) 均能过，但都忘了，结果只打了个暴力。难受。板子题，题解就不放了

Code

#include<bits/stdc++.h>
#define ri register int
#define p(i) ++i
using namespace std;
namespace IO{
    char buf[1<<21],*p1=buf,*p2=buf;
    #define gc() p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++
    inline int read() {
        ri x=0,f=1;char ch=getchar();
        while(ch<\'0\'||ch>\'9\') {if (ch==\'-\') f=-1;ch=getchar();}
        while(ch>=\'0\'&&ch<=\'9\') {x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
        return x*f; 
    }
}
using IO::read;
namespace nanfeng{
    #define cmax(x,y) ((x)>(y)?(x):(y))
    #define cmin(x,y) ((x)>(y)?(y):(x))
    #define FI FILE *IN
    #define FO FILE *OUT
    typedef unsigned long long ull;
    static const int N=1e5+7;
    static const ull P=131;
    char a[N<<1],b[N];
    int st[N],cnt,T,la,lb,ans;
    ull ah[N],bh[N],p[N];
    inline int check(int x) {
        if (ah[x]==bh[lb]-bh[lb-x]*p[x]) return 1;
        return 0;
    }
    inline void init() {cnt=0;ans=0;}
    inline int main() {
        // FI=freopen("nanfeng.in","r",stdin);
        // FO=freopen("nanfeng.out","w",stdout);
        T=read();
        p[0]=1;
        for (ri i(1);i<=N-5;p(i)) p[i]=p[i-1]*P;
        for (ri z(1);z<=T;p(z)) {
            init();
            la=read(),lb=read();
            scanf("%s",a+1);
            scanf("%s",b+lb+1);
            for (ri i(1);i<=lb;p(i)) {
                if (i<=la&&a[i]==b[lb+1]) st[p(cnt)]=i; 
                b[i]=a[i];
                bh[i]=ah[i]=ah[i-1]*P+(ull)(a[i]);
            }
            lb+=1;
            if (lb<=la&&a[lb]==b[lb]) {printf("%d\\n",ans=lb);continue;}
            bh[lb]=bh[lb-1]*P+(ull)(b[lb]);
            for (ri i(cnt);i;--i) if (check(st[i])) {ans=st[i];break;}
            printf("%d\\n",ans);
        }
        return 0;
    }
}
int main() {return nanfeng::main();}

T2

一道 \\(tarjan\\) 缩点的题，结果光求了割点，没缩，还乐呵了半天

题解

T3

一道很好的利用单调性的题，可以通过维护一个单调指针，答案递推而得

题解