HDU 5351 MZL's Border

HDU 5351 MZL’s Border

题意

定义字符串f1=b，f2=a，fi=fi−1fi−2。
对fn的长度为m的前缀s，求最大的k满足s[1]=s[m−k+1],s[2]=s[m−k+2]...s[k]=s[m]

Solution

先列出前几个字符串
f3=ab,f4=aba,f5=abaab,f6=abaababa,f7=abaababaabaab。
对于n，m。
假设fn−1的长度大于等于m，那么相当于求n-1，m。
假设fn−1的小于m
由斐波那契的性质，答案要么是m−length(fn−1),
要么是m−length(fn−2)
通过对前几个字符串的分析可以发现，当m+1<length(fn)时答案是后者。
于是只要先把前1000个字符串的长度求出来。需要使用高精度。
然后直接按照上面的规则判断就行了。

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#include <bits/stdc++.h>#define lson x << 1#define rson x << 1 | 1#define LL long long#define se second#define fi first#define LL long longusing namespace std;const int N = 1003;string  fib[N], m, ans;inline string add ( string a, string b ){    if ( a.size() < b.size() ) swap ( a, b );    int la = a.size(), lb = b.size(), x = 0, i = 0;    for ( i = 0; i < lb; ++i ) {        x += ( a[i] - '0' + b[i] - '0' );        a[i] = '0' + x % 10;        x /= 10;    }    while ( x && i < la ) {        x += ( a[i] - '0' );        a[i] = '0' + x % 10;        x /= 10;        ++i;    }    if ( x ) a += '0' + x;    return a;}inline string sub ( string a, string b ){    int la = a.size(), lb = b.size(), x = 0, i = 0;    for ( i = 0; i < lb; ++i ) {        if ( a[i] - x >= b[i] ) {            a[i] = a[i] - x - b[i] + '0';            x = 0;        } else {            a[i] = 10 + '0' + a[i] - b[i] - x;            x = 1;        }    }    while ( x && i < la ) {        if ( a[i] != '0' ) {            a[i] = a[i] - 1;            x = 0;        } else {            a[i] = '9';            ++i;        }    }    i = la - 1;    while ( a[i] == '0' && i != 0 ) {        a.erase ( a.begin() + i );        --i;    }    return a;}int T, n;inline bool pd ( string a, string b ){    if ( a.size() == b.size() )  {        for ( int i = a.size() - 1; i >= 0; --i ) {            if ( a[i] != b[i] ) return a[i] > b[i];        }        return 0;    }    return a.size() > b.size();}void make ( int n, string m ){    if ( n < 4 ) {        ans = "0";        return ;    }    if ( pd ( m, fib[n - 1] ) ) {        string s1 = sub ( fib[n], "2" );        if ( !pd ( m, s1 ) ) {            ans = sub ( m, fib[n - 2] );        } else {            ans = sub ( m, fib[n - 1] );;        }        return ;    } else make ( n - 1, m );}void MOD ( string s, LL k ){    LL tem = 0;    for ( int i = s.size() - 1; i >= 0; --i ) {        tem = ( tem * 10 + s[i] - '0' ) % k;    }    cout << tem << endl;}int main(){    fib[0] = "0";    fib[1] = "1";    for ( int i = 2; i <= 1000; ++i ) {        fib[i] = add ( fib[i - 1] , fib[i - 2] );    }    cin >> T;    while ( T-- ) {        cin >> n >> m;        reverse ( m.begin(), m.end() );        make ( n, m );        MOD ( ans, 258280327 );    }}