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  • 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

      1 // BEGIN CUT HERE
  2 /*
  3  * Author:  plum rain
  4  * score :
  5  */
  6 /*
  7 
  8  */
  9 // END CUT HERE
 10 #line 11 "RevealTriangle.cpp"
 11 #include <sstream>
 12 #include <stdexcept>
 13 #include <functional>
 14 #include <iomanip>
 15 #include <numeric>
 16 #include <fstream>
 17 #include <cctype>
 18 #include <iostream>
 19 #include <cstdio>
 20 #include <vector>
 21 #include <cstring>
 22 #include <cmath>
 23 #include <algorithm>
 24 #include <cstdlib>
 25 #include <set>
 26 #include <queue>
 27 #include <bitset>
 28 #include <list>
 29 #include <string>
 30 #include <utility>
 31 #include <map>
 32 #include <ctime>
 33 #include <stack>
 34 
 35 using namespace std;
 36 
 37 #define CLR(x) memset(x, 0, sizeof(x))
 38 #define CLR1(x) memset(x, -1, sizeof(x))
 39 #define PB push_back
 40 #define SZ(v) ((int)(v).size())
 41 #define ALL(t) t.begin(),t.end()
 42 #define zero(x) (((x)>0?(x):-(x))<eps)
 43 #define out(x) cout<<#x<<":"<<(x)<<endl
 44 #define tst(a) cout<<#a<<endl
 45 #define CINBEQUICKER std::ios::sync_with_stdio(false)
 46 
 47 typedef vector<int> VI;
 48 typedef vector<string> VS;
 49 typedef vector<double> VD;
 50 typedef pair<int, int> pii;
 51 typedef long long int64;
 52 
 53 const double eps = 1e-8;
 54 const double PI = atan(1.0)*4;
 55 const int maxint = 2139062143;
 56 
 57 class RevealTriangle
 58 {
 59     public:
 60         vector <string> calcTriangle(vector <string> q){
 61             int sz = q.size();
 62             string s;
 63             for (int i = sz-2; i >= 0; -- i){
 64                 s = q[i];
 65                 int pos = 0;
 66                 for (int j = 0; j < s.size(); ++ j)
 67                     if (s[j] >= '0' && s[j] <= '9'){
 68                         pos = j; break;
 69                     }
 70 
 71                 for (int k = pos-1; k >= 0; -- k)
 72                     for (int t = 0; t < 10; ++ t)
 73                         if ((t+s[k+1]-'0') % 10 == q[i+1][k] - '0')
 74                             s[k] = t + '0';
 75                 for (int k = pos+1; k < s.size(); ++ k)
 76                     for (int t = 0; t < 10; ++ t)
 77                         if ((t+s[k-1]-'0') % 10 == q[i+1][k-1] - '0')
 78                             s[k] = t + '0';
 79 
 80                 q[i] = s;
 81             }
 82             return q;
 83         }
 84 
 85         // BEGIN CUT HERE
 86     public:
 87         void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0(); if ((Case == -1) || (Case == 1)) test_case_1(); if ((Case == -1) || (Case == 2)) test_case_2(); if ((Case == -1) || (Case == 3)) test_case_3(); }
 88     private:
 89         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(); }
 90         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 << "	Expected: " << print_array(Expected) << endl; cerr << "	Received: " << print_array(Received) << endl; } }
 91         void test_case_0() { string Arr0[] = {"4??", 
 92             "?2", 
 93             "1"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); string Arr1[] = {"457", "92", "1" }; vector <string> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); verify_case(0, Arg1, calcTriangle(Arg0)); }
 94         void test_case_1() { string Arr0[] = {"1"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); string Arr1[] = {"1" }; vector <string> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); verify_case(1, Arg1, calcTriangle(Arg0)); }
 95         void test_case_2() { string Arr0[] = {"???2", "??2", "?2", "2"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); string Arr1[] = {"0002", "002", "02", "2" }; vector <string> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); verify_case(2, Arg1, calcTriangle(Arg0)); }
 96         void test_case_3() { string Arr0[] = {"??5?", "??9", "?4", "6"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); string Arr1[] = {"7054", "759", "24", "6" }; vector <string> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); verify_case(3, Arg1, calcTriangle(Arg0)); }
 97 
 98         // END CUT HERE
 99 
100 };
101 
102 // BEGIN CUT HERE
103 int main()
104 {
105     //    freopen( "a.out" , "w" , stdout );    
106     RevealTriangle ___test;
107     ___test.run_test(-1);
108     return 0;
109 }
110 // END CUT HERE
    View Code

    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

      1 // BEGIN CUT HERE
  2 /*
  3 
  4 */
  5 // END CUT HERE
  6 #line 7 "KSubstring.cpp"
  7 #include <cstdlib>
  8 #include <cctype>
  9 #include <cstring>
 10 #include <cstdio>
 11 #include <cmath>
 12 #include <algorithm>
 13 #include <vector>
 14 #include <iostream>
 15 #include <sstream>
 16 #include <set>
 17 #include <queue>
 18 #include <fstream>
 19 #include <numeric>
 20 #include <iomanip>
 21 #include <bitset>
 22 #include <list>
 23 #include <stdexcept>
 24 #include <functional>
 25 #include <string>
 26 #include <utility>
 27 #include <map>
 28 #include <ctime>
 29 #include <stack>
 30 
 31 using namespace std;
 32 
 33 #define CLR(x) memset(x, 0, sizeof(x))
 34 #define PB push_back
 35 #define SZ(v) ((int)(v).size())
 36 #define ALL(x) (x).begin(), (x).end()
 37 #define out(x) cout<<#x<<":"<<(x)<<endl
 38 #define tst(a) cout<<#a<<endl
 39 #define CINBEQUICKER std::ios::sync_with_stdio(false)
 40 
 41 typedef vector<int> VI;
 42 typedef vector<string> VS;
 43 typedef vector<double> VD;
 44 typedef long long int64;
 45 
 46 const double eps = 1e-8;
 47 const double PI = atan(1.0)*4;
 48 const int maxint = 2139062143;
 49 
 50 inline int MyMod( int a , int b ) { return (a%b+b)%b;}
 51 
 52 struct nod{
 53     int64 num, s;
 54 };
 55 
 56 int64 a[5000];
 57 nod num[5000];
 58 int pos[5000];
 59 
 60 bool cmp(nod a, nod b)
 61 {
 62     return b.num < a.num || (a.num == b.num && a.s < b.s);
 63 }
 64 
 65 class KSubstring
 66 {
 67     public:
 68         vector <int> maxSubstring(int a0, int x, int y, int mod, int n){
 69             a[0] = a0;
 70             for (int i = 1; i < n; ++ i) a[i] = (a[i-1]*x + y) % mod;
 71 
 72             int64 val = 1LL<<62;
 73             int k_idx = 0;
 74             for (int k = 1; k <= n/2; ++ k){
 75                 int64 sum = 0, idx = 0;
 76                 for (int i = 0; i < k; ++ i) sum += a[i];
 77                 for (int i = 0; i+k <= n; ++ i){
 78                     num[idx].num = sum;
 79                     num[idx++].s = i;
 80                     sum = sum - a[i] + a[i+k];
 81                 }
 82 
 83                 sort(num, num+idx, cmp);
 84                 int p = idx-1;
 85                 int64 tnum = num[idx-1].num;
 86                 for (int i = idx-1; i >= 0; -- i){
 87                     if (num[i].num != tnum){
 88                         pos[i] = i+1; p = i; tnum = num[i].num;
 89                     }
 90                     else pos[i] = p;
 91                 }
 92                 for (int i = 0; i < idx; ++ i){
 93                     int64 tmp = abs(num[i].num - num[pos[i]].num);
 94                     int64 pdif = abs(num[pos[i]].s - num[i].s);
 95                     if (pdif >= k && val >= tmp){
 96                         k_idx = k; val = tmp;
 97                     }
 98                 }
 99             }
100             VI ans; ans.PB (k_idx); ans.PB ((int)val);
101             return ans;
102         }
103         
104 // BEGIN CUT HERE
105     public:
106     //void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_2();}
107     void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0(); if ((Case == -1) || (Case == 1)) test_case_1(); if ((Case == -1) || (Case == 2)) test_case_2(); if ((Case == -1) || (Case == 3)) test_case_3(); }
108     private:
109     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(); }
110     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 << "	Expected: " << print_array(Expected) << endl; cerr << "	Received: " << print_array(Received) << endl; } }
111     void test_case_0() { int Arg0 = 5; int Arg1 = 3; int Arg2 = 4; int Arg3 = 25; int Arg4 = 5; int Arr5[] = {2, 1 }; vector <int> Arg5(Arr5, Arr5 + (sizeof(Arr5) / sizeof(Arr5[0]))); verify_case(0, Arg5, maxSubstring(Arg0, Arg1, Arg2, Arg3, Arg4)); }
112     void test_case_1() { int Arg0 = 8; int Arg1 = 19; int Arg2 = 17; int Arg3 = 2093; int Arg4 = 12; int Arr5[] = {5, 161 }; vector <int> Arg5(Arr5, Arr5 + (sizeof(Arr5) / sizeof(Arr5[0]))); verify_case(1, Arg5, maxSubstring(Arg0, Arg1, Arg2, Arg3, Arg4)); }
113     void test_case_2() { int Arg0 = 53; int Arg1 = 13; int Arg2 = 9; int Arg3 = 65535; int Arg4 = 500; int Arr5[] = {244, 0 }; vector <int> Arg5(Arr5, Arr5 + (sizeof(Arr5) / sizeof(Arr5[0]))); verify_case(2, Arg5, maxSubstring(Arg0, Arg1, Arg2, Arg3, Arg4)); }
114     void test_case_3() { int Arg0 = 12; int Arg1 = 34; int Arg2 = 55; int Arg3 = 7890; int Arg4 = 123; int Arr5[] = {35, 4 }; vector <int> Arg5(Arr5, Arr5 + (sizeof(Arr5) / sizeof(Arr5[0]))); verify_case(3, Arg5, maxSubstring(Arg0, Arg1, Arg2, Arg3, Arg4)); }
115 
116 // END CUT HERE
117 
118 };
119 //by plum rain
120 // BEGIN CUT HERE
121 int main()
122 {
123     //freopen( "a.out" , "w" , stdout );    
124     KSubstring ___test;
125     ___test.run_test(-1);
126        return 0;
127 }
128 // END CUT HERE
    View Code
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  • 原文地址：https://www.cnblogs.com/plumrain/p/SRM_404.html
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