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.
#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 <bits/stdc++.h> #include <algorithm> #include <map> #include <queue> #include <set> #include <stack> #include <string> #include <vector> 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<<ans;**/ n=read; dfs(0,0); cout<<ans; end; }
