HDU 6148 Valley Numer (数位DP)

题意：。。。

析：好久没写数位DP了，几乎就是不会了。。。。

dp[i][last][s] 表示前 i 位上一位是 last，当前的状态是 s，0表示非上升，1 表示非下降，然后就很简单了，只有 0 能转成 1，1就是最后的状态。

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 <bitset>
#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 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 = 1e16;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1000 + 10;
const LL mod = 1000000007;
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[maxn][11][2];
char s[maxn];
int a[maxn];

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

  int n = ok ? a[pos] : 9;
  LL res = 0;
  if(last == 10){
    res += dfs(pos-1, 10, s, 1, ok && 0 == n);
    for(int i = 1; i <= n; ++i)
      res += dfs(pos-1, i, 0, 0, i == n && ok);
  }
  else if(s){
    for(int i = last; i <= n; ++i)
      res += dfs(pos-1, i, s, i == 0 && is, ok && i == n);
  }
  else{
    for(int i = 0; i <= n; ++i)
      if(i > last)  res += dfs(pos-1, i, 1, is && i == 0, ok && i == n);
      else  res += dfs(pos-1, i, 0, is && i == 0, ok && i == n);
  }
  res %= mod;
  if(!is && !ok)  ans = res;
  return res;
}

LL solve(char *s){
  int len = 0;
  n = strlen(s);
  for(int i = n-1; i >= 0; --i)
    a[++len] = s[i] - '0';
  return dfs(len, 10, 0, 1, 1);
}

int main(){
  ms(dp, -1);
  int T;  cin >> T;
  while(T--){
    scanf("%s", s);
    printf("%I64d\n", solve(s));
  }
  return 0;
}