hdu 6156 palindrome function #数位dp 3

数位dp ， 注意‘】pos < (len + 1 )/2 以及每次模k

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#include<cstdio>#include<cstring>#include<iostream>#include<string>using namespace std;typedef long long ll ;const int maxn = 10005 ;int temp[100] ;ll dp[40][100][100][2] ;int a[20] ;int b[20] ;int all ;ll dfs( int pos ,   int len ,bool status , bool limit ,int k ){    if( pos == -1  )    return status ? k : 1 ;    if( !limit && dp[k][pos][len][status] != -1 ) return dp[k][pos][len][status] ;    int up = limit ? a[pos] : k - 1 ;    ll tmp = 0 ;    for( int i = 0 ; i <= up ; i ++ ){        temp[pos] = i ;        if( i == 0 && pos == len )            tmp += dfs( pos-1 , len-1 , status , limit && i == up , k ) ;        else if( pos < (len + 1 )/2   )            tmp += dfs( pos-1 , len , status && i == temp[len-pos] , limit && i == up , k ) ;        else            tmp += dfs( pos-1 , len , status , limit && i == up , k ) ;    }    if(! limit) dp[k][pos][len][status] = tmp ;    return tmp ;}ll solve (ll L ,  int k) {    int po = 0 ;    while( L ){        a[po ++ ] = L % k ;        L /= k ;    }    int t = 0 ;    return dfs( po - 1 , po - 1  , true , true , k ) ;}int main () {    int kase , ka = 1 ;    ll L , R , l , r ;    scanf("%d",&kase);    memset(dp,-1,sizeof dp);    while( kase -- )    {        scanf("%lld %lld %lld %lld",&L,&R , &l , &r);        ll res = 0 ;        for( int i = l ; i <= r  ; i ++ ){            res += solve( R , i ) - solve( L-1 , i );        }        printf("Case #%d: %lld\n",ka++, res );    }    return 0 ;}