HDU 6156 Palindrome Function 数位DP

  题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=6156

  题目描述: 求L~R所有的数的l~r进制的f(x), f(x) = 当前进制 如果回文串, f(x) = 1 其他情况

  解题思路: 数位DP, 统计个数 , 需要作出改变的就是进制和回文

  代码: 

#include <iostream>
#include <cstdio>
#include <string>
#include <vector>
#include <cstring>
#include <iterator>
#include <cmath>
#include <algorithm>
#include <stack>
#include <deque>
#include <map>
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
#define mem0(a) memset(a,0,sizeof(a))
#define mem1(a) memset(a,-1,sizeof(a))
#define meminf(a) memset(a,0x3f,sizeof(a))
#define fi(n) for(i=0;i<n;i++)
#define fj(m) for(j=0;j<m;j++)
#define sca(x) scanf("%d",&x)
#define scalld(x) scanf("%I64d",&x)
#define print(x) printf("%d
",x)
#define printlld(x) printf("%I64d
",x)
#define d printf("=======
")

typedef long long LL;
using namespace std;
const int maxn = 100;

int digit[maxn];
int temp[maxn];

LL dp[40][maxn][maxn][5]; // dp[i][j][p][q] i 进制第j位是p完成还是没完成

LL dfs( int k, int pos, int len, int status, int limit ) {
    if( pos < 0 ) {
        if( status ) return k;
        else return 1;
    }
    if( !limit && dp[k][pos][len][status] != -1 ) {
        return dp[k][pos][len][status];
    }
    int end = limit ? digit[pos] : k-1;
    LL ret = 0;
    for( int i = 0; i <= end; i++ ) {
        temp[pos] = i;
        if( i == 0 && len == pos) {
            ret += dfs( k, pos-1, len-1, status, limit && (i==end));
        }
        else if( status && pos < (len+1)/2 ) {
            ret += dfs( k, pos-1, len, i == temp[len-pos], limit&&(i==end) );
        }
        else {
            ret += dfs( k, pos-1, len, status, limit&&(i==end) );
        }
    }
    if( !limit ) dp[k][pos][len][status] = ret;
    return ret;
}


LL fun( int num, int k ) {
    if( num == 0 ) return k;
    int cnt = 0;
    while( num ) {
        digit[cnt++] = num % k;
        num /= k;
    }
    return dfs( k, cnt-1, cnt-1, 1, 1 );
}

int main() {
    int t;
//    cin >> t;
    sca(t);
    int cases = 1;
    mem1(dp);
    while( t-- ) {
        LL res = 0;
        int L, R, l, r;
//        cin >> L >> R >> l >> r;
        sca(L), sca(R), sca(l), sca(r);
        for( int i = l; i <= r; i++ ) {
            res += fun(R, i) - fun(L-1, i);
        }
        printf( "Case #%d: %lld
", cases++, res );
//        cout << "Case: #" << cases++ << ": " << res << endl;
    }
    return 0;
}
View Code

  思考: 自己一开始没想到数位DP, 赛后补题的时候写的时候也有很多错误, 有好多细节自己想不到, 是真的菜, 现在把代码删了, 重新写一遍, 还有好多好多地方是需要努力的啊

http://acm.hdu.edu.cn/showproblem.php?pid=6156

转载于:https://www.cnblogs.com/FriskyPuppy/p/7398339.html

原文地址:https://www.cnblogs.com/twodog/p/12139595.html