SRM 403(1-250pt, 1-500pt)

DIV1 250pt

题意：称各个数位只含有4和7的数为lucky number，给定a，b，求[a, b]中的lucky number有多少个。a, b <= 10^9

解法：很明显的数位dp，写出来也就20行左右。本来以为最近刚好在做数位dp，可以很快出，结果我想着刚才做的那个数位dp，脑袋进水居然直接敲了段递归的代码还半天没调出来.......然后很快写了段非递归的代码过了.....160score...

　　　后来看官方题解说，由于10^9内实际上只有1022个lucky number，所以可以求全部生成出来，看有哪些在[a, b]上。

tag:数位dp，think

 // BEGIN CUT HERE
 /*
  * Author:  plum rain
  * score :
  */
 /*

  */
 // END CUT HERE
 #line 11 "TheLuckyNumbers.cpp"
 #include <sstream>
 #include <stdexcept>
 #include <functional>
 #include <iomanip>
 #include <numeric>
 #include <fstream>
 #include <cctype>
 #include <iostream>
 #include <cstdio>
 #include <vector>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <cstdlib>
 #include <set>
 #include <queue>
 #include <bitset>
 #include <list>
 #include <string>
 #include <utility>
 #include <map>
 #include <ctime>
 #include <stack>

 using namespace std;

 #define CLR(x) memset(x, 0, sizeof(x))
 #define CLR1(x) memset(x, -1, sizeof(x))
 #define PB push_back
 #define SZ(v) ((int)(v).size())
 #define ALL(t) t.begin(),t.end()
 #define zero(x) (((x)>0?(x):-(x))<eps)
 #define out(x) cout<<#x<<":"<<(x)<<endl
 #define tst(a) cout<<#a<<endl
 #define CINBEQUICKER std::ios::sync_with_stdio(false)

 typedef vector<int> VI;
 typedef vector<string> VS;
 typedef vector<double> VD;
 typedef pair<int, int> pii;
 typedef long long int64;

 const double eps = 1e-;
 const double PI = atan(1.0)*;
 const int maxint = ;

 int dit[], two[];

 int gao(int x)
 {
     two[] = ;
     for (int i = ; i < ; ++ i)
         two[i] = two[i-] * ;

     int len = ;
     while (x){
         dit[len++] = x % ;
         x /= ;
     }

     int ret = ;
     for (int i = ; i < len; ++ i)
         ret += two[i];
     for (int i = len-; i >= ; -- i){
         if ( < dit[i]) ret += two[i];
         if ( < dit[i]) ret += two[i];
         if (dit[i] !=  && dit[i] != ) break;
     }
     return ret;
 }

 class TheLuckyNumbers
 {
     public:
         int count(int a, int b){
             return gao(b+) - gao(a);
         }

 // BEGIN CUT HERE
     public:
     void run_test(int Case) { if ((Case == -) || (Case == )) test_case_0(); if ((Case == -) || (Case == )) test_case_1(); if ((Case == -) || (Case == )) test_case_2(); if ((Case == -) || (Case == )) test_case_3(); }
     private:
     template <typename T> string print_array(const vector<T> &V) { ostringstream os; os << "{ "; for (typename vector<T>::const_iterator iter = V.begin(); iter != V.end(); ++iter) os << '\"' << *iter << "\","; os << " }"; return os.str(); }
     void verify_case(int Case, const int &Expected, const int &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << Received << '\"' << endl; } }
     void test_case_0() { int Arg0 = ; int Arg1 = ; int Arg2 = ; verify_case(, Arg2, count(Arg0, Arg1)); }
     void test_case_1() { int Arg0 = ; int Arg1 = ; int Arg2 = ; verify_case(, Arg2, count(Arg0, Arg1)); }
     void test_case_2() { int Arg0 = ; int Arg1 = ; int Arg2 = ; verify_case(, Arg2, count(Arg0, Arg1)); }
     void test_case_3() { int Arg0 = ; int Arg1 = ; int Arg2 = ; verify_case(, Arg2, count(Arg0, Arg1)); }

 // END CUT HERE

 };

 // BEGIN CUT HERE
 int main()
 {
 //    freopen( "a.out" , "w" , stdout );
     TheLuckyNumbers ___test;
     ___test.run_test(-);
        return ;
 }
 // END CUT HERE

DIV1 500pt

题意：称各个数位只含有4和7的数为lucky number，称完全由来自vector<int> numbers(含有lucky number和也含有unlucky number)里面的lucky number组成，且每一个a[i]的末位数字和a[i+1]的首位数字均相同的序列为lucky sequence。给定vector<int> numbers和int n，求长度为n的lucky sequence有多少个。只要一个数不同，则lucky sequence视为不同。

解法：lucky number分为四种，分别为4开头4结尾的，4开头7结尾的，7开头4结尾的，7开头7结尾的。若a[i]属于第A类，且a[i+1]可以为B类，则从A向B连接一条有向边。这样，就相当于求长度为n的路径有多少条。

　　　我最初的想法是二分，设f(x, s, e)表示求长度为x，从s类开始，从e类结束的路径有多少条，f(x, s, e)由 f(x/2, s, i) 和 f(x-x/2, i, e)求得。我以为这样二分以后时间复杂度降低了，实际上每次二分都有4个决策，时间复杂度反而提升了，于是我毫无悬念地TLE了。然后，我改了代码，把x / 2变成了x / 10000，即将x分为10001段来求，由于前面每一段长度均为x/10000，后面一段长度为x - x/10000，所以实际上只需要求20个东西，这样时间复杂度就降低到能接受的范围了。

　　　虽然算是水过了这道题，还是蛮开心的.....

　　　官方题解给出的方法是用矩阵快速幂优化。其实要构造这个并不难吧。。。当时想了一下用矩阵，怎么就没想出来呢。。。。

tag:矩阵乘法

 // BEGIN CUT HERE
 /*
  * Author:  plum rain
  * score :
  */
 /*

  */
 // END CUT HERE
 #line 11 "TheLuckySequence.cpp"
 #include <sstream>
 #include <stdexcept>
 #include <functional>
 #include <iomanip>
 #include <numeric>
 #include <fstream>
 #include <cctype>
 #include <iostream>
 #include <cstdio>
 #include <vector>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <cstdlib>
 #include <set>
 #include <queue>
 #include <bitset>
 #include <list>
 #include <string>
 #include <utility>
 #include <map>
 #include <ctime>
 #include <stack>

 using namespace std;

 #define CLR(x) memset(x, 0, sizeof(x))
 #define CLR1(x) memset(x, -1, sizeof(x))
 #define PB push_back
 #define SZ(v) ((int)(v).size())
 #define ALL(t) t.begin(),t.end()
 #define zero(x) (((x)>0?(x):-(x))<eps)
 #define out(x) cout<<#x<<":"<<(x)<<endl
 #define tst(a) cout<<#a<<endl
 #define CINBEQUICKER std::ios::sync_with_stdio(false)

 typedef vector<int> VI;
 typedef vector<string> VS;
 typedef vector<double> VD;
 typedef pair<int, int> pii;
 typedef long long int64;

 const double eps = 1e-;
 const double PI = atan(1.0)*;
 const int maxint = ;
 const int mod = ;

 int64 cnt[];
 pii temp[];
 set<int> st;
 int64 pat[][], d[][][];
 int64 an[][];

 int gao(int x)
 {
     pii num;
     num.second = x % ;
     while (x >= ){
         int tmp = x % ;
         x /= ;
         if (tmp !=  && tmp != ) return -;
     }
     num.first = x % ;
     for (int i = ; i < ; ++ i)
         if (num == temp[i]) return i;
     return -;
 }

 int64 rec(int len, int s, int e)
 {
     if (!len) return s == e;
     if (len == ) return pat[s][e];

     if (len <  && d[len][s][e] != -)
         return d[len][s][e];

     if (len < ){
         int haf = len / ;
         int64 &ret = d[len][s][e];
         ret = ;
         for (int i = ; i < ; ++ i)
             ret = (ret + rec(haf, s, i) * rec(len-haf, i, e)) % mod;
         return ret;
     }

     int haf = len / ;
     CLR (an);
     for (int i = ; i < ; ++ i)
         an[][i] = rec(haf, s, i);
     for (int i = ; i < ; ++ i)
         for (int j = ; j < ; ++ j)
             for (int k = ; k < ; ++ k)
                 an[i][j] = (an[i][j] + an[i-][k] * rec(haf, k, j)) % mod;
     int64 ret = ;
     for (int i = ; i < ; ++ i)
         ret = (ret + an[][i] * rec(len-haf*, i, e)) % mod;
     return ret;
 }

 class TheLuckySequence
 {
     public:
         int count(vector <int> num, int len){
             temp[] = make_pair(,); temp[] = make_pair(,);
             temp[] = make_pair(,); temp[] = make_pair(,);

             CLR (cnt); st.clear();
             for (int i = ; i < (int)num.size(); ++ i){
                 int tmp = gao(num[i]);
                 if (tmp != - && !st.count(num[i])){
                     ++ cnt[tmp];
                     st.insert(num[i]);
                 }
             }
             CLR (pat);
             for (int i = ; i < ; ++ i)
                 for (int j = ; j < ; ++ j)
                     if (temp[i].second == temp[j].first){
                         pat[i][j] = cnt[i];
                     }

             CLR1 (d);
             int64 ret = ;
             for (int i = ; i < ; ++ i)
                 for (int j = ; j < ; ++ j)
                     ret = (ret + rec(len-, i, j) * cnt[j]) % mod;
             return (int)ret;
         }

 // BEGIN CUT HERE
     public:
     void run_test(int Case) { if ((Case == -) || (Case == )) test_case_0(); if ((Case == -) || (Case == )) test_case_1(); if ((Case == -) || (Case == )) test_case_2(); if ((Case == -) || (Case == )) test_case_3(); }
     private:
     template <typename T> string print_array(const vector<T> &V) { ostringstream os; os << "{ "; for (typename vector<T>::const_iterator iter = V.begin(); iter != V.end(); ++iter) os << '\"' << *iter << "\","; os << " }"; return os.str(); }
     void verify_case(int Case, const int &Expected, const int &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << Received << '\"' << endl; } }
     void test_case_0() { int Arr0[] = {, , , , , , , }; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; int Arg2 = ; verify_case(, Arg2, count(Arg0, Arg1)); }
     void test_case_1() { int Arr0[] = {, , }; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; int Arg2 = ; verify_case(, Arg2, count(Arg0, Arg1)); }
     void test_case_2() { int Arr0[] = {, , , , }; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; int Arg2 = ; verify_case(, Arg2, count(Arg0, Arg1)); }
     void test_case_3() { int Arr0[] = {, , , }; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; int Arg2 = ; verify_case(, Arg2, count(Arg0, Arg1)); }

 // END CUT HERE

 };

 // BEGIN CUT HERE
 int main()
 {
 //    freopen( "a.out" , "w" , stdout );
     TheLuckySequence ___test;
     ___test.run_test(-);
        return ;
 }
 // END CUT HERE