SRM 404(1-250pt, 1-500pt)

DIV1 250pt

题意：对于1-9数字三角形如下图，设其为a[i][j]，则a[i][j] = (a[i-1][j] + a[i-1][j+1]) % 10。现在对于某个数字三角形， 每行告诉你某一个位置的数字，求整个数字三角形。

　　　n <= 50。

解法：首先，由于最后一行和倒数第二行共3个数，已知2个，所以就能求出剩下一个，然后一个往上，一行一行将每个数求出来。时间复杂度小于O(10*n^2)。

tag:think

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

  */
 // END CUT HERE
 #line 11 "RevealTriangle.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 = ;

 class RevealTriangle
 {
     public:
         vector <string> calcTriangle(vector <string> q){
             int sz = q.size();
             string s;
             for (int i = sz-; i >= ; -- i){
                 s = q[i];
                 int pos = ;
                 for (int j = ; j < s.size(); ++ j)
                     if (s[j] >= '' && s[j] <= ''){
                         pos = j; break;
                     }

                 for (int k = pos-; k >= ; -- k)
                     for (int t = ; t < ; ++ t)
                         if ((t+s[k+]-'') %  == q[i+][k] - '')
                             s[k] = t + '';
                 for (int k = pos+; k < s.size(); ++ k)
                     for (int t = ; t < ; ++ t)
                         if ((t+s[k-]-'') %  == q[i+][k-] - '')
                             s[k] = t + '';

                 q[i] = s;
             }
             return q;
         }

         // 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 vector <string> &Expected, const vector <string> &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "\tExpected: " << print_array(Expected) << endl; cerr << "\tReceived: " << print_array(Received) << endl; } }
         void test_case_0() { string Arr0[] = {"4??",
             "?2",
             ""}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); string Arr1[] = {"", "", "" }; vector <string> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); verify_case(, Arg1, calcTriangle(Arg0)); }
         void test_case_1() { string Arr0[] = {""}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); string Arr1[] = {"" }; vector <string> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); verify_case(, Arg1, calcTriangle(Arg0)); }
         void test_case_2() { string Arr0[] = {"???2", "??2", "?2", ""}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); string Arr1[] = {"", "", "", "" }; vector <string> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); verify_case(, Arg1, calcTriangle(Arg0)); }
         void test_case_3() { string Arr0[] = {"??5?", "??9", "?4", ""}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); string Arr1[] = {"", "", "", "" }; vector <string> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); verify_case(, Arg1, calcTriangle(Arg0)); }

         // END CUT HERE

 };

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

DIV1 500pt

题意：通过O(n)的方法生成一个长度为n的数列a[]。定义s(i,k) = a[i] + a[i+1] + ... + a[i+k-1]。求最小的|s(i,k) - s(j,k)|，其中i+k-1 < j，并返回k和|s(i,k) - s(j,k)|。如果有多组解，返回k值较大的。

解法：枚举k，求出k固定时最小的满足题意的i, j, k。下面考虑k已经固定的情况。

　　　法一：维护一个set，然后用二分查找绝对值差最小的一组。这样的方法时间复杂度O(n*n*logn)，但是常数很大，不一定能过。

　　　法二：考虑一个结论，对于任意一组i, j, k，如果i+k-1 >= j，则一定存在另一组i', j', k'满足i' + k' - 1 < j'，且|s(i, k) - s(j, k)| = |s(i', k') - s(j', k')|。

　　　所以得到的方法是，首先记录答案返回值为k_idx 和 val，对于所有的s(i, j)做一排序，第一影响因素是s(i, j)从小到大，第二影响因素是i从小到大，将排序结果存在num[]中。对于num[]中的相邻元素t1和t2(t1在t2之前)，有两种可能：

　　　1、如果t1和t2的s(i, k)之差不为0，则if(|t1.s - t2.s| <= val && |t1.i - t2.i| >= k) 更新答案val和k_idx，否则不更新。

　　　2、如果t1和t2的s(i, k)相等，则考虑找出所有s(i, k)与他们相同的项，然后取i之差最大的两项来比较。当然，这个要提前预处理好，不然会很慢。具体见代码。

tag:think, good

 // BEGIN CUT HERE
 /*

 */
 // END CUT HERE
 #line 7 "KSubstring.cpp"
 #include <cstdlib>
 #include <cctype>
 #include <cstring>
 #include <cstdio>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 #include <iostream>
 #include <sstream>
 #include <set>
 #include <queue>
 #include <fstream>
 #include <numeric>
 #include <iomanip>
 #include <bitset>
 #include <list>
 #include <stdexcept>
 #include <functional>
 #include <string>
 #include <utility>
 #include <map>
 #include <ctime>
 #include <stack>

 using namespace std;

 #define CLR(x) memset(x, 0, sizeof(x))
 #define PB push_back
 #define SZ(v) ((int)(v).size())
 #define ALL(x) (x).begin(), (x).end()
 #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 long long int64;

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

 inline int MyMod( int a , int b ) { return (a%b+b)%b;}

 struct nod{
     int64 num, s;
 };

 int64 a[];
 nod num[];
 int pos[];

 bool cmp(nod a, nod b)
 {
     return b.num < a.num || (a.num == b.num && a.s < b.s);
 }

 class KSubstring
 {
     public:
         vector <int> maxSubstring(int a0, int x, int y, int mod, int n){
             a[] = a0;
             for (int i = ; i < n; ++ i) a[i] = (a[i-]*x + y) % mod;

             int64 val = 1LL<<;
             int k_idx = ;
             for (int k = ; k <= n/; ++ k){
                 int64 sum = , idx = ;
                 for (int i = ; i < k; ++ i) sum += a[i];
                 for (int i = ; i+k <= n; ++ i){
                     num[idx].num = sum;
                     num[idx++].s = i;
                     sum = sum - a[i] + a[i+k];
                 }

                 sort(num, num+idx, cmp);
                 int p = idx-;
                 int64 tnum = num[idx-].num;
                 for (int i = idx-; i >= ; -- i){
                     if (num[i].num != tnum){
                         pos[i] = i+; p = i; tnum = num[i].num;
                     }
                     else pos[i] = p;
                 }
                 for (int i = ; i < idx; ++ i){
                     int64 tmp = abs(num[i].num - num[pos[i]].num);
                     int64 pdif = abs(num[pos[i]].s - num[i].s);
                     if (pdif >= k && val >= tmp){
                         k_idx = k; val = tmp;
                     }
                 }
             }
             VI ans; ans.PB (k_idx); ans.PB ((int)val);
             return ans;
         }

 // BEGIN CUT HERE
     public:
     //void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_2();}
     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 vector <int> &Expected, const vector <int> &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "\tExpected: " << print_array(Expected) << endl; cerr << "\tReceived: " << print_array(Received) << endl; } }
     void test_case_0() { int Arg0 = ; int Arg1 = ; int Arg2 = ; int Arg3 = ; int Arg4 = ; int Arr5[] = {,  }; vector <int> Arg5(Arr5, Arr5 + (sizeof(Arr5) / sizeof(Arr5[]))); verify_case(, Arg5, maxSubstring(Arg0, Arg1, Arg2, Arg3, Arg4)); }
     void test_case_1() { int Arg0 = ; int Arg1 = ; int Arg2 = ; int Arg3 = ; int Arg4 = ; int Arr5[] = {,  }; vector <int> Arg5(Arr5, Arr5 + (sizeof(Arr5) / sizeof(Arr5[]))); verify_case(, Arg5, maxSubstring(Arg0, Arg1, Arg2, Arg3, Arg4)); }
     void test_case_2() { int Arg0 = ; int Arg1 = ; int Arg2 = ; int Arg3 = ; int Arg4 = ; int Arr5[] = {,  }; vector <int> Arg5(Arr5, Arr5 + (sizeof(Arr5) / sizeof(Arr5[]))); verify_case(, Arg5, maxSubstring(Arg0, Arg1, Arg2, Arg3, Arg4)); }
     void test_case_3() { int Arg0 = ; int Arg1 = ; int Arg2 = ; int Arg3 = ; int Arg4 = ; int Arr5[] = {,  }; vector <int> Arg5(Arr5, Arr5 + (sizeof(Arr5) / sizeof(Arr5[]))); verify_case(, Arg5, maxSubstring(Arg0, Arg1, Arg2, Arg3, Arg4)); }

 // END CUT HERE

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