제출 #437889

#제출 시각아이디문제언어결과실행 시간메모리
437889HaidaraDNA 돌연변이 (IOI21_dna)C++17
0 / 100
67 ms6464 KiB
    /* * * * * * * * * *
     *   ID: Haidara   *
     *   LANG: C++17   *
     *   PROB:         *
     * * * * * * * * * */
    #include<bits/stdc++.h>
    #define fast ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
     
    #define rep(i,x,n) for(int i=x;i<n;i++)
    #define FOR(i,n) rep(i,0,n)
    #define per(i,x,n) for(int i=x;i>n;i--)
    #define ROF(i,x) for(int i=x;i>=0;i--)
    #define v(i) vector< i >
    #define p(i,j) pair< i , j >
    #define pii pair<int,int>
    #define m(i,j) map< i , j >
    #define pq(i) priority_queue< i >
    #define ff first
    #define all(x) x.begin(),x.end()
    #define ss second
    #define pp push_back
    using namespace std;
    const int mod=1e9+7;
    const int maxn=100100;
    int st[3][2][maxn*4];
    string str[2];
    int sz;
    int idx;
    void build(int l=1,int r=sz,int inx=1)
    {
        if(l==r)
        {
            if(str[idx][l]=='T')
                st[2][idx][inx]=1;
            if(str[idx][l]=='C')
                st[2][idx][inx]=1;
            if(str[idx][l]=='A')
                st[2][idx][inx]=1;
            return ;
        }
        int mid=l+(r-l)/2;
        build(l,mid,inx*2);
        build(mid+1,r,inx*2+1);
        FOR(i,3)
        st[i][idx][inx]=st[i][idx][inx*2]+st[i][idx][inx*2+1];
        return ;
    }
    int f[3],s[3];
    void query(int ql,int qr,int l=1,int r=sz,int inx=1)
    {
        if(ql<=l&&r<=qr)
        {
            FOR(i,3)
            f[i]+=st[i][0][inx],s[i]+=st[i][1][inx];
            return ;
        }
        if(l>qr||r<ql)
            return ;
        int mid=l+(r-l)/2;
        query(ql,qr,l,mid,inx*2);
        query(ql,qr,mid+1,r,inx*2+1);
        return ;
    }
    void init(string a, string b)
    {
        str[0]=a;
        str[1]=b;
      	sz=max(a.size(),b.size());
      idx=0;
      build();
      idx=1;
      build();
    }
    int get_distance(int x, int y)
    {
        x++,y++;
        FOR(i,3)
        f[i]=s[i]=0;
        int ans=y-x;
        query(x,y);
        FOR(i,3)
        if(f[i]!=s[i])
            ans=-1;
        return ans;
    }
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