HDU 6148 Valley Numer （数位DP）题解

思路：

只要把status那里写清楚就没什么难度T^T，当然还要考虑前导零！

代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
using namespace std;
const int N = 100000+5;
const int MOD = 1000000007;
int dp[110][15][5]; //不知0，升1，降2，降升3
int a[N];
ll dfs(ll pos,int pre,int sta,bool limit,bool lead){
    if(pos == -1) return lead? 0 : 1;
    if(!limit && dp[pos][pre][sta] != -1) return dp[pos][pre][sta];
    int top = limit? a[pos] : 9;
    ll ret = 0;
    int sta2;
    for(int i = 0;i <= top;i++){
        if(sta == 0){
            if(lead) sta2 = 0;
            else if(i > pre) sta2 = 1;
            else if(i < pre) sta2 = 2;
            else sta2 = 0;
        }
        else if(sta == 1){
            if(i < pre) continue;
            else sta2 = 1;
        }
        else if(sta == 2){
            if(i > pre) sta2 = 3;
            else sta2 = 2;
        }
        else if(sta == 3){
            if(i >= pre) sta2 = 3;
            else continue;
        }
        ret += dfs(pos-1,i,sta2,limit && i == top,lead && i == 0);
        ret %= MOD;
    }
    ret %= MOD;
    if(!limit && !lead) dp[pos][pre][sta] = ret;

    return ret;
}
int main(){
    int T;
    char s[105];
    memset(dp,-1,sizeof(dp));
    scanf("%d",&T);
    while(T--){
        scanf("%s",s);
        int len = strlen(s),pos = 0;
        for(int i = len - 1;i >=0 ;i--){
            a[pos++] = s[i] - '0';
        }
        printf("%lld\n",dfs(pos-1,0,0,true,true));
    }
    return 0;
}