刚开始发现小于 n 位的二级制形式数为 2^n-1 个，然后再在剩下的区间处理判断，发现TLE ,于是只好 DFS。
One beautiful July morning a terrible thing happened in Mainframe: a mean virus Megabyte somehow got access to the memory of his not less mean sister Hexadecimal. He loaded there a huge amount of n different natural numbers from 1 to n to obtain total control over her energy.
But his plan failed. The reason for this was very simple: Hexadecimal didn't perceive any information, apart from numbers written in binary format. This means that if a number in a decimal representation contained characters apart from 0 and 1, it was not stored in the memory. Now Megabyte wants to know, how many numbers were loaded successfully.
Input
Input data contains the only number n (1 ≤ n ≤ 109).
Output
Output the only number — answer to the problem.
Examples
input
Copy
10
output
Copy
2
Note
For n = 10 the answer includes numbers 1 and 10.
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
#include<set>
#include<map>
using namespace std;
typedef long long ll;
#define maxn 2000005
#define inf 0x3f3f3f3f
#define ii 0x3f
const int mod = 1e9 + 7;
// dp[i]=min(dp[j]+(i-j-1+sum[i]-sum[j]-l)^2)
// dp[i] 表示前 i 个玩具装箱的费用
// f[i]=sum[i]+i;
// c=1+l;
// f[i] 单增
ll read() {
ll x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-') {
f = -1;
}
ch = getchar();
}
while (ch >= '0'&&ch <= '9') {
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}
ll quickpow(ll a, ll b) {
ll ans = 1;
while (b > 0) {
if (b % 2)ans = ans * a;
b = b / 2;
a = a * a;
}
return ans;
}
//char s[maxn];
int ans;
ll n;
void dfs(ll x) {
if (x > n)return;
ans++;
dfs(x * 10);
dfs(x * 10 + 1);
}
int main() {
ios::sync_with_stdio(false);
//ll n;
cin >> n;
ans = 0;
dfs(1);
cout << ans << endl;
}
刚开始发现小于 n 位的二级制形式数为 2^n-1 个，然后再在剩下的区间处理判断，发现TLE ,于是只好 DFS。
One beautiful July morning a terrible thing happened in Mainframe: a mean virus Megabyte somehow got access to the memory of his not less mean sister Hexadecimal. He loaded there a huge amount of n different natural numbers from 1 to n to obtain total control over her energy.
But his plan failed. The reason for this was very simple: Hexadecimal didn't perceive any information, apart from numbers written in binary format. This means that if a number in a decimal representation contained characters apart from 0 and 1, it was not stored in the memory. Now Megabyte wants to know, how many numbers were loaded successfully.
Input
Input data contains the only number n (1 ≤ n ≤ 109).
Output
Output the only number — answer to the problem.
Examples
input
Copy
10
output
Copy
2
Note
For n = 10 the answer includes numbers 1 and 10.
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
#include<set>
#include<map>
using namespace std;
typedef long long ll;
#define maxn 2000005
#define inf 0x3f3f3f3f
#define ii 0x3f
const int mod = 1e9 + 7;
// dp[i]=min(dp[j]+(i-j-1+sum[i]-sum[j]-l)^2)
// dp[i] 表示前 i 个玩具装箱的费用
// f[i]=sum[i]+i;
// c=1+l;
// f[i] 单增
ll read() {
ll x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-') {
f = -1;
}
ch = getchar();
}
while (ch >= '0'&&ch <= '9') {
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}
ll quickpow(ll a, ll b) {
ll ans = 1;
while (b > 0) {
if (b % 2)ans = ans * a;
b = b / 2;
a = a * a;
}
return ans;
}
//char s[maxn];
int ans;
ll n;
void dfs(ll x) {
if (x > n)return;
ans++;
dfs(x * 10);
dfs(x * 10 + 1);
}
int main() {
ios::sync_with_stdio(false);
//ll n;
cin >> n;
ans = 0;
dfs(1);
cout << ans << endl;
}
