数位dp一道自动机题 poj 3208

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <algorithm>
#include <string.h>
using namespace std;
#define maxn 20
#define LL long long
LL digit[maxn],dp[maxn][10],n;
 
 
LL dfs (LL pos ,LL have ,bool doing){
    if(pos == -1) return have == 3;
    if(!doing && dp[pos][have]!=-1)
        return dp[pos][have];
    LL ans = 0 ,nhave ,npre;
    LL end = doing ? digit[pos]:9;
 
    for(int i=0;i <= end;i ++){
        if(have == 3) nhave = 3;
        else if(have == 2 && i==6) nhave = 3;
        else if(have == 1 && i==6) nhave = 2;
        else if(have == 0 && i==6) nhave = 1;
        else nhave = 0;
        ans += dfs(pos - 1 , nhave , doing && (i == end));
    }
    if(!doing) dp[pos][have] = ans;
    return ans;
}
LL calc(LL x){
    memset(dp,-1,sizeof(dp));
    LL pos = 0;
    while(x){
        digit[pos ++] = x %10;
        x /= 10;
    }
    return dfs(pos - 1, 0 ,1);
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%I64d",&n);
        LL l = 666LL,mid;
        LL r = 100000000000LL;
        while(l < r){
            mid = (l + r)/2;
            LL temp = calc(mid);
            if(temp >= n) r = mid;
            else l = mid + 1;
        }
        printf("%I64d\n",l);
    }
    return 0;
}

用二分找出符合题目要求的值,然后用dp找出那个小于等于那个值有多少个含3个666的字符串,从而找出答案,dfs为数位dp的常用形式

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转载自blog.csdn.net/qq_42193011/article/details/87388074