UVaLive 3357 Pinary (Fib数列+递归)

题意：求第 k 个不含前导 0 和连续 1 的二进制串。

析：1，10，100，101，1000，...很容易发现长度为 i 的二进制串的个数正好就是Fib数列的第 i 个数，因为第 i 个也有子问题，其子问题也就是Fib，这样就可以用递归来解决了。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#define debug() puts("++++")
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define be begin()
#define ed end()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
//#define all 1,n,1
#define FOR(i,n,x)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.in", "r", stdin)
#define freopenw freopen("out.out", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int maxn = 1000 + 10;
const int maxm = 1e5 + 10;
const LL mod = 1000000007;
const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};
const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}
inline int readInt(){ int x;  scanf("%d", &x);  return x; }
vector<int> v;

void dfs(int n, int last){
  if(n == 0){
    while(last-- > 0)  putchar('0');
    return ;
  }
  int pos = lower_bound(v.be, v.ed, n) - v.be;
  if(n < v[pos])  --pos;
  for(int i = pos; i < last; ++i)  putchar('0');
  putchar('1');
  dfs(n-v[pos], pos-1);
}

int main(){
  v.pb(1);  v.pb(1);
  for(int i = 2; i < 40; ++i)  v.pb(v[i-1] + v[i-2]);
  v[0] = 0;
  int T;  cin >> T;
  while(T--){
    scanf("%d", &n);
    dfs(n, -1);
    putchar('\n');
  }
  return 0;
}