CodeForces 1166D Cute Sequences

题目链接：http://codeforces.com/problemset/problem/1166/D

题目大意

　　给定序列的第一个元素 a 和最后一个元素 b 还有一个限制 m，请构造一个序列，序列的第 i 项 x_i 满足 $x_i = \sum_{i = 1}^{i - 1} + r_i，1 \leq r_i \leq m$。

分析

　　令 r_i 分别全 1 和全 m，可以得到$2^{i - 2} * (a + 1) \leq x_i \leq 2^{i - 2} * (a + m), i \geq 2$。

　　因此，如果以 b 为第 k 个元素能够构成一个序列，那么 b 一定满足$2^{k - 2} * (a + 1) \leq b \leq 2^{k - 2} * (a + m), k \geq 2$。

　　反过来，如果存在 k 使 b 满足$2^{k - 2} * (a + 1) \leq b \leq 2^{k - 2} * (a + m), k \geq 2$，那么是否一定能构成一个序列呢？其实是可以的，证明如下：

$$
\begin{align*}
&∵\ x_i = (\sum_{j = 1}^{j - 1} x_j) + r_i，1 \leq r_i \leq m，i \geq 2 \\
&∵\ \sum_{j = 1}^{j - 1} x_j = 2 * x_{i - 1} - r_{i - 1} \\
&∴\ x_i = 2 * x_{i - 1} + r_i - r_{i - 1} \\
&∵\ x_2 = a + r_2 \\
&∴\ x_i = 2^{i - 2} * a + 2^{i - 3} * r_2 + 2^{i - 4} * r_3 + \dots + 2 * r_{i - 2} + r_{i - 1} + r_i \\
&∴\ x_i = 2^{i - 2} * a + r_i + (\sum_{j = 0}^{i - 3} 2^j * r_{i - j - 1}) \\
\ &设 h = r_k，d_i = r_k - h，1 - m \leq d_i \leq m - 1，1 \leq h \leq m \\
&∴\ x_i = 2^{i - 2} * a + h + (\sum_{j = 0}^{i - 3} 2^j * (d_{i - j - 1} + h)) \\
&∴\ x_i = 2^{i - 2} * (a + h) + (\sum_{j = 0}^{i - 3} 2^j * d_{i - j - 1}) \\
\ &令 d_{i - j - 1} = 0\ or \ 1，0 \leq j \leq i - 3 \\
\ &于是有(\sum_{j = 0}^{i - 3} 2^j * d_{i - j - 1}) < 2^{i - 2} \\
&∴\ x_i = 2^{i - 2} * (a + h) + (\sum_{j = 0}^{i - 3} 2^j * d_{i - j - 1}) 满足 2^{i - 2} * (a + 1) \leq x_i \leq 2^{i - 2} * (a + m), i \geq 2
\end{align*}
$$

　　至此我们找到了 $r_i$ 的一种赋值方案，使得 $2^{i - 2} * (a + 1) \leq x_i \leq 2^{i - 2} * (a + m), i \geq 2$。

　　应用到通项公式后我们发现： $x_{i - 1} = \frac{x_i + r_{i - 1} - r_i}{2} = \frac{x_i + p_i}{2}，p_i \in \{-1, 0, 1\}$。

　　p_i 的值视 x_i 而定，且满足$2 \ | \ (x_i + p_i)$。

　　那么序列 p 中把 0 除外后的序列必然是 -1，1，-1交错的，可以用反证法证明。

　　如果 p_i == p_i+1 == 1，那么 r_i-1 - r_i+1 == 2，与 |r_j - r_i| == |d_j - d_i| <= 1 矛盾。

　　据此可以从尾部开始向前构造序列。

代码如下

 #include <bits/stdc++.h>
 using namespace std;

 #define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
 #define Rep(i,n) for (int i = 0; i < (n); ++i)
 #define For(i,s,t) for (int i = (s); i <= (t); ++i)
 #define rFor(i,t,s) for (int i = (t); i >= (s); --i)
 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)

 #define pr(x) cout << #x << " = " << x << "  "
 #define prln(x) cout << #x << " = " << x << endl

 #define LOWBIT(x) ((x)&(-x))

 #define ALL(x) x.begin(),x.end()
 #define INS(x) inserter(x,x.begin())

 #define ms0(a) memset(a,0,sizeof(a))
 #define msI(a) memset(a,inf,sizeof(a))
 #define msM(a) memset(a,-1,sizeof(a))

 #define MP make_pair
 #define PB push_back
 #define ft first
 #define sd second

 template<typename T1, typename T2>
 istream &operator>>(istream &in, pair<T1, T2> &p) {
     in >> p.first >> p.second;
     return in;
 }

 template<typename T>
 istream &operator>>(istream &in, vector<T> &v) {
     for (auto &x: v)
         in >> x;
     return in;
 }

 template<typename T1, typename T2>
 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
     out << "[" << p.first << ", " << p.second << "]" << "\n";
     return out;
 }

 inline int gc(){
     static const int BUF = 1e7;
     static char buf[BUF], *bg = buf + BUF, *ed = bg;

     if(bg == ed) fread(bg = buf, , BUF, stdin);
     return *bg++;
 } 

 inline int ri(){
     int x = , f = , c = gc();
     for(; c<||c>; f = c=='-'?-:f, c=gc());
     for(; c>&&c<; x = x* + c - , c=gc());
     return x*f;
 }

 typedef long long LL;
 typedef unsigned long long uLL;
 typedef pair< double, double > PDD;
 typedef pair< int, int > PII;
 typedef pair< string, int > PSI;
 typedef set< int > SI;
 typedef vector< int > VI;
 typedef map< int, int > MII;
 typedef pair< LL, LL > PLL;
 typedef vector< LL > VL;
 typedef vector< VL > VVL;
 const double EPS = 1e-;
 const LL inf = 0x7fffffff;
 const LL infLL = 0x7fffffffffffffffLL;
 const LL mod = 1e18 + ;
 const int maxN = 2e5 + ;
 const LL ONE = ;
 const LL evenBits = 0xaaaaaaaaaaaaaaaa;
 const LL oddBits = 0x5555555555555555;

 int q;
 LL a, b, m, k;
 LL x[];

 int main(){
     INIT();
     cin >> q;
     while(q--) {
         cin >> a >> b >> m;
         LL p = ;
         k = ;
         // 找使得 b <= p * (a + m) 的最小 p
         while(b > p * (a + m)) {
             ++k;
             p <<= ;
         }
         // 在 a != b 的情况下，只要满足 p * (a + 1) <= b <= p * (a + m)，就一定可以构造
         if(b >= p * (a + )) {
             cout << k << " ";
             x[] = a;
             x[k] = b;
             int p = ;
             rFor(i, k - , ) {
                 if(x[i + ] %  == ) x[i] = x[i + ] >> ;
                 else {
                     x[i] = (x[i + ] + p) >> ;
                     p = -p;
                 }
             }
             For(i, , k) cout << x[i] << " " << endl;
             cout << endl;
         }
         else if(a == b) cout << "1 " << a << endl;
         else cout << - << endl;
     }
     return ;
 }

大佬代码如下

　　和我的代码效果是一样的，但我死活推不出原理，只能先搁着里了。

 #include <bits/stdc++.h>
 using namespace std;

 #define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
 #define Rep(i,n) for (int i = 0; i < (n); ++i)
 #define For(i,s,t) for (int i = (s); i <= (t); ++i)
 #define rFor(i,t,s) for (int i = (t); i >= (s); --i)
 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)

 #define pr(x) cout << #x << " = " << x << "  "
 #define prln(x) cout << #x << " = " << x << endl

 #define LOWBIT(x) ((x)&(-x))

 #define ALL(x) x.begin(),x.end()
 #define INS(x) inserter(x,x.begin())

 #define ms0(a) memset(a,0,sizeof(a))
 #define msI(a) memset(a,inf,sizeof(a))
 #define msM(a) memset(a,-1,sizeof(a))

 #define MP make_pair
 #define PB push_back
 #define ft first
 #define sd second

 template<typename T1, typename T2>
 istream &operator>>(istream &in, pair<T1, T2> &p) {
     in >> p.first >> p.second;
     return in;
 }

 template<typename T>
 istream &operator>>(istream &in, vector<T> &v) {
     for (auto &x: v)
         in >> x;
     return in;
 }

 template<typename T1, typename T2>
 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
     out << "[" << p.first << ", " << p.second << "]" << "\n";
     return out;
 }

 inline int gc(){
     static const int BUF = 1e7;
     static char buf[BUF], *bg = buf + BUF, *ed = bg;

     if(bg == ed) fread(bg = buf, , BUF, stdin);
     return *bg++;
 } 

 inline int ri(){
     int x = , f = , c = gc();
     for(; c<||c>; f = c=='-'?-:f, c=gc());
     for(; c>&&c<; x = x* + c - , c=gc());
     return x*f;
 }

 typedef long long LL;
 typedef unsigned long long uLL;
 typedef pair< double, double > PDD;
 typedef pair< int, int > PII;
 typedef pair< string, int > PSI;
 typedef set< int > SI;
 typedef vector< int > VI;
 typedef map< int, int > MII;
 typedef pair< LL, LL > PLL;
 typedef vector< LL > VL;
 typedef vector< VL > VVL;
 const double EPS = 1e-;
 const LL inf = 0x7fffffff;
 const LL infLL = 0x7fffffffffffffffLL;
 const LL mod = 1e18 + ;
 const int maxN = 2e5 + ;
 const LL ONE = ;
 const LL evenBits = 0xaaaaaaaaaaaaaaaa;
 const LL oddBits = 0x5555555555555555;

 int q;
 LL a, b, m, k;

 int main(){
     INIT();
     cin >> q;
     while(q--) {
         cin >> a >> b >> m;
         LL p = ;
         k = ;
         // 找使得 b <= p * (a + m) 的最小 p
         while(b > p * (a + m)) {
             ++k;
             p <<= ;
         }
         // 在 a != b 的情况下，只要满足 p * (a + 1) <= b <= p * (a + m)，就一定可以构造
         if(b >= p * (a + )) {
             cout << k << " " << a << " ";
             For(i, , k - ) cout << (((b >> (k - i - )) + ) >> ) << " ";
             cout << b << endl;
         }
         else if(a == b) cout << "1 " << a << endl;
         else cout << - << endl;
     }
     return ;
 }