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BZOJ 1026 [SCOI2009]windy数 (数位DP)

1026: [SCOI2009]windy数

Time Limit: 1 Sec  Memory Limit: 162 MB
Submit: 8753  Solved: 3950
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Description

  windy定义了一种windy数。不含前导零且相邻两个数字之差至少为2的正整数被称为windy数。 windy想知道,
在A和B之间,包括A和B,总共有多少个windy数?

Input

  包含两个整数,A B。

Output

  一个整数

Sample Input

【输入样例一】
1 10
【输入样例二】
25 50

Sample Output

【输出样例一】
9
【输出样例二】
20

HINT

 

【数据规模和约定】

100%的数据,满足 1 <= A <= B <= 2000000000 。

 

Source

析:dp[pos][last] 表示前 pos 位上一位是last,很简单的一个数位DP。

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>
#include <bitset>
#include <numeric>
#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 cl clear()
//#define all 1,n,1
#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)
#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 LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 400 + 10;
const int maxm = 3e5 + 10;
const ULL mod = 3;
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;
}

int dp[20][20];
int a[20];

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

  int res = 0;
  int n = ok ? a[pos] : 9;
  for(int i = 0; i <= n; ++i)
    if(!i && is)  res += dfs(pos-1, last, 1, i == n && ok);
    else if(abs(i - last) > 1)  res += dfs(pos-1, i, 0, i == n && ok);
  if(!is && !ok)  ans = res;
  return res;
}

int solve(int n){
  int len = 0;
  while(n){
    a[++len] = n % 10;
    n /= 10;
  }
  return dfs(len, 11, 1, 1);
}

int main(){
  scanf("%d %d", &m, &n);  ms(dp, -1);
  printf("%d\n", solve(n) - solve(m-1));
  return 0;
}

  

posted on 2017-11-15 21:26  dwtfukgv  阅读(...)  评论(... 编辑 收藏