AtCoder--755——dfs

题目描述

You are given an integer N. Among the integers between 1 and N (inclusive), how many Shichi-Go-San numbers (literally “Seven-Five-Three numbers”) are there?

Here, a Shichi-Go-San number is a positive integer that satisfies the following condition:

When the number is written in base ten, each of the digits 7, 5 and 3 appears at least once, and the other digits never appear.
Constraints
1≤N<109
N is an integer.

输入

Input is given from Standard Input in the following format:

N

输出

Print the number of the Shichi-Go-San numbers between 1 and N (inclusive).

样例输入

575

样例输出

4

提示

There are four Shichi-Go-San numbers not greater than 575: 357,375,537 and 573.

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#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize (2)
#pragma G++ optimize (2)
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
#define wuyt main
typedef long long ll;
#define HEAP(...) priority_queue<__VA_ARGS__ >
#define heap(...) priority_queue<__VA_ARGS__,vector<__VA_ARGS__ >,greater<__VA_ARGS__ > >
template<class T> inline T min(T &x,const T &y){
     return x>y?y:x;}
template<class T> inline T max(T &x,const T &y){
     return x<y?y:x;}
//#define getchar()(p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
//char buf[(1 << 21) + 1], *p1 = buf, *p2 = buf;
ll read(){
     ll c = getchar(),Nig = 1,x = 0;while(!isdigit(c) && c!='-')c = getchar();
if(c == '-')Nig = -1,c = getchar();
while(isdigit(c))x = ((x<<1) + (x<<3)) + (c^'0'),c = getchar();
return Nig*x;}
#define read read()
const ll inf = 1e15;
const int maxn = 2e5 + 7;
const int mod = 1e9 + 7;
#define start int wuyt()
#define end return 0
int n,ans;
void dfs(int sum, unsigned char ss){
     
    if(sum>n)
        return;
    if(ss==07)
        ans++;
    if(sum<99999999){
     
        dfs(sum*10+3,ss|1);
        dfs(sum*10+5,ss|2);
        dfs(sum*10+7,ss|4);
    }
}
start{
     
    /**
    ll x=read;
    if(x==7||x==5||x==3)
        printf("YES\n");
    else
        printf("NO\n");
    string s;
    cin>>s;
    int len=s.size();
    int ans=999;
    for(int i=0;i<=len-3;++i)
    {
        int num=(s[i]-'0')*100+(s[i+1]-'0')*10+(s[i+2]-'0');
        ans=min(abs(num-753),ans);
    }
    cout<
    n=read;
    dfs(0,0);
    cout<<ans;
    end;
}