HDU 6156 Palindrome Function (数位DP)

题意：给定上 l，r，a， b，让你求从 l 到 r 这个区间内的在a进制到b进制如果是回文数就加上该进制的大小，如果不是，那么就加上1.

析：数位DP。

dp[i][j][k] 表示 i 进制下，前 j 位，回文串的长度是k有多少个，然后统计完后，再乘以进制即可，最后再加上那些不是回文。

但是好像跑的有点慢，3700ms多，不知道什么情况。。。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 63248 + 100;
const int mod = 1e9;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
    return r > 0 && r <= n && c > 0 && c <= m;
}

LL dp[40][40][40];
int num[40];
int a[40];

LL dfs(int x, int pos, int last, bool is, bool ok){
  if(!pos)  return 1;
  LL &ans = dp[x][pos][last];
  if(!is && !ok && ans >= 0)  return ans;

  LL res = 0;
  int n = ok ? a[pos] : x-1;
  for(int i = 0; i <= n; ++i){
    if(is){
      num[last] = i;
      res += dfs(x, pos-1, last-(i == 0), i == 0, i == n && ok);
    }
    else{
      int m = (last+1) / 2;
      if(last&1){
        if(pos >= m){
          num[pos] = i;
          res += dfs(x, pos-1, last, is && i == 0, i == n && ok);
        }
        else if(i == num[2*m-pos])
          res += dfs(x, pos-1, last, is && i == 0, ok && i == n);
      }
      else{
        if(pos > m){
          num[pos] = i;
          res += dfs(x, pos-1, last, is && i == 0, i == n && ok);
        }
        else if(i == num[last+1-pos])
          res += dfs(x, pos-1, last, is && i == 0, i == n && ok);
      }
    }
  }

  if(!ok && !is)  ans = res;
  return res;
}

LL solve(int n, int x, int y){
  LL ans = 0;
  LL sum = 0;
  for(int i = x; i <= y; ++i){
    int xx = n;
    int len = 0;
    while(xx){
      a[++len] = xx % i;
      xx /= i;
    }
    LL tmp = dfs(i, len, len, 1, 1);
    sum += tmp;
    ans += tmp * i;
  }
  ans += (LL)(y-x+1) * n - sum;
  return ans;
}

inline bool scan(int &ret){
  char c;
  if(c=getchar(),c==EOF)  return 0;
  while(c<'0'&&c>'9')c=getchar();
  ret=c-'0';
  while(c=getchar(),c>='0'&&c<='9')ret=ret*10+(c-'0');
  return true;
}

int main(){
  ms(dp, -1);
  int T;  scan(T);
  for(int kase = 1; kase <= T; ++kase){
    int a, b, l, r;
    scan(l); scan(r);  scan(a);  scan(b);
    printf("Case #%d: %I64d
", kase, solve(r, a, b) - solve(l-1, a, b));
  }
  return 0;
}