SRM 406(1-250pt, 1-500pt)

DIV1 250pt

题意：有几家宠物店，vecort<int>A表示每家宠物店含有小狗占小狗总数的百分比。现在要做扇形统计图统计每家店的小狗百分比，如下图，问作出来的扇形统计图中最多含有多少对半径夹角为180度。(左图两对，右图一对) (A.size() <= 8)

解法：因为A.size() <= 8，所以直接暴力枚举A中元素的全排列就好了。我的代码又写复杂了。。。一是枚举全排列可以用next_permutation()函数，另一个是对每个排列统计数量的时候写复杂了，不需要算j < i的情况，只要不除以2就行了。

tag:brute-force

  1 // BEGIN CUT HERE
  2 /*
  3  * Author:  plum rain
  4  * score :
  5  */
  6 /*
  7 
  8  */
  9 // END CUT HERE
 10 #line 11 "SymmetricPie.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 clr0(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<<" "
 45 #define tst1(a) cout<<#a<<endl
 46 #define CINBEQUICKER std::ios::sync_with_stdio(false)
 47 
 48 typedef vector<int> VI;
 49 typedef vector<string> VS;
 50 typedef vector<double> VD;
 51 typedef pair<int, int> pii;
 52 typedef long long int64;
 53 
 54 const double eps = 1e-8;
 55 const double PI = atan(1.0)*4;
 56 const int inf = 2139062143 / 2;
 57 
 58 int n, ans;
 59 VI an, dn;
 60 bool mp[10];
 61 int sum[10];
 62 
 63 void dfs(int x)
 64 {
 65     if (x == n){
 66         clr0 (sum); sum[0] = dn[an[0]];
 67         for (int i = 1; i < n; ++ i)
 68             sum[i] = sum[i-1] + dn[an[i]];
 69 
 70         int cnt = 0;
 71         for (int i = 0; i < n; ++ i)
 72             for (int j = 0; j < n; ++ j){
 73                 int tmp = j > i ? sum[j] - sum[i] : sum[n-1] - sum[i] + sum[j];
 74                 if (tmp == 50) ++ cnt;
 75             }
 76         ans = max(ans, cnt/2);
 77         return ;
 78     }
 79 
 80     for (int i = 0; i < n; ++ i) if (!mp[i]){
 81         an.pb (i); mp[i] = 1;
 82         dfs (x + 1); 
 83         an.pop_back(); mp[i] = 0;
 84     }
 85 }
 86 
 87 class SymmetricPie
 88 {
 89     public:
 90         int getLines(vector <int> D){
 91             dn = D;
 92             ans = 0; n = sz(dn);
 93             clr0 (mp); an.clear();
 94             dfs (0);
 95             return ans;
 96         }
 97         
 98 // BEGIN CUT HERE
 99     public:
100     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(); if ((Case == -1) || (Case == 4)) test_case_4(); }
101     private:
102     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(); }
103     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 << "	Expected: "" << Expected << '"' << endl; cerr << "	Received: "" << Received << '"' << endl; } }
104     void test_case_0() { int Arr0[] = {10,40,10,40}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 2; verify_case(0, Arg1, getLines(Arg0)); }
105     void test_case_1() { int Arr0[] = {10,50,40}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 1; verify_case(1, Arg1, getLines(Arg0)); }
106     void test_case_2() { int Arr0[] = {50,50}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 1; verify_case(2, Arg1, getLines(Arg0)); }
107     void test_case_3() { int Arr0[] = {1,48,1,1,48,1}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 3; verify_case(3, Arg1, getLines(Arg0)); }
108     void test_case_4() { int Arr0[] = {2,2,96}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 0; verify_case(4, Arg1, getLines(Arg0)); }
109 
110 // END CUT HERE
111 
112 };
113 
114 // BEGIN CUT HERE
115 int main()
116 {
117 //    freopen( "a.out" , "w" , stdout );    
118     SymmetricPie ___test;
119     ___test.run_test(-1);
120        return 0;
121 }
122 // END CUT HERE
View Code

DIV1 500pt

题意：有一张纸，纸上有n * m格，每个格子上有一个数字。若将纸折叠(可以不对折，但是折痕只能在格与格之间，不能穿过格子。)，折叠后重合的两个格子原先的值之和即为现在的值，比如1*2的纸含有两个数字1,2，折叠后含有一个数字3。折叠完成后，最终纸上所有数字中最大的为max。问随意折叠多少次，求max最大为多少。n <= 12，m <= 12，-100 <= A[i][j] <= 100

解法：也是暴力，只不过要预处理+暴力。只是开始我想到暴力的时候觉得这样不好写，而且感觉好慢可能不行，就没有细想了。。。所以也没有做出来。暴力的方法是，先预处理哪些列最终可能折叠在一起，再预处理哪些行最终可能会折叠在一起，然后对于处理出来的东西枚举就好了。我看了官方题解也没想好怎么写，然后又看了它推荐的代码才弄懂。

tag:brute-force

  1 // BEGIN CUT HERE
  2 /*
  3  * Author:  plum rain
  4  * score :
  5  */
  6 /*
  7 
  8  */
  9 // END CUT HERE
 10 #line 11 "FoldThePaper.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 clr0(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<<" "
 45 //#define tst1(a) cout<<#a<<endl
 46 #define CINBEQUICKER std::ios::sync_with_stdio(false)
 47 
 48 typedef vector<int> vi;
 49 typedef vector<string> vs;
 50 typedef vector<double> vd;
 51 typedef pair<int, int> pii;
 52 typedef long long int64;
 53 
 54 const double eps = 1e-8;
 55 const double PI = atan(1.0)*4;
 56 const int inf = 2139062143 / 2;
 57 
 58 int n, m;
 59 int an[55][55];
 60 bool can[1<<12];
 61 vector<vi> row, col;
 62 
 63 vi getvec(int x, int f)
 64 {
 65     vi v;
 66     for (int i = 0; i < (f ? m : n); ++ i)
 67         if (x & (1<<i)) v.pb (i);
 68     return v;
 69 }
 70 
 71 void rec(vi v){
 72     //for (int i = 0; i < sz(v); ++ i)
 73         //tst (v[i]);
 74     //cout << endl;
 75     int n = sz(v);
 76     for (int i = 0; i < n; ++ i)
 77         can[v[i]] = 1;
 78     for (int i = 1; i < n; ++ i){
 79         vi nv(max(i, n-i));
 80         if (i <= n/2){
 81             for (int j = i; j < n; ++ j)
 82                 nv[j-i] = v[j];
 83             for (int j = 0, k = i-1; j < i; ++ j, -- k)
 84                 nv[j] |= v[k];
 85         }
 86         else{
 87             for (int j = 0; j < i; ++ j)
 88                 nv[j] = v[j];
 89             for (int j = i, k = i-1; j < n; ++ j, -- k)
 90                 nv[k] |= v[j];
 91         }
 92         rec(nv);
 93     }
 94 }
 95 
 96 class FoldThePaper
 97 {
 98     public:
 99         int getValue(vector <string> pap){
100             n = sz(pap);
101             stringstream stm;
102             for (int i = 0; i < n; ++ i){
103                 stm << pap[i]; m = 0;
104                 while (stm >> an[i][m]) ++ m;
105                 stm.clear();
106             }
107             clr0 (can);
108             vi v;
109             for (int i = 0; i < n; ++ i) v.pb (1<<i);
110             rec(v);
111             row.clear();
112             for (int i = 0; i < (1<<n); ++ i) if (can[i])
113                 row.pb (getvec(i, 0));
114 
115             v.clear(); clr0 (can);
116             for (int i = 0; i < m; ++ i) v.pb (1<<i);
117             rec(v);
118             col.clear();
119             for (int i = 0; i < (1<<m); ++ i) if (can[i])
120                 col.pb (getvec(i, 1));
121 
122             int ans = -inf;
123             for (int i = 0; i < sz(row); ++ i)
124                 for (int j = 0; j < sz(col); ++ j){
125                     int tmp = -inf;
126                     for (int t = 0; t < sz(row[i]); ++ t)
127                         for (int k = 0; k < sz(col[j]); ++ k){
128                             if (tmp == -inf) tmp = an[row[i][t]][col[j][k]];
129                             else tmp += an[row[i][t]][col[j][k]];
130                         }
131                     ans = max(tmp, ans);
132                 }
133             return ans;
134         }
135         
136 // BEGIN CUT HERE
137     public:
138     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(); if ((Case == -1) || (Case == 4)) test_case_4(); }
139     private:
140     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(); }
141     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 << "	Expected: "" << Expected << '"' << endl; cerr << "	Received: "" << Received << '"' << endl; } }
142     void test_case_0() { string Arr0[] = {
143 "1 1 1",
144 "1 1 1"
145 }; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 6; verify_case(0, Arg1, getValue(Arg0)); }
146     void test_case_1() { string Arr0[] = {
147 "1 -1",
148 "1 -1"
149 }; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 2; verify_case(1, Arg1, getValue(Arg0)); }
150     void test_case_2() { string Arr0[] = {
151 "1 -1 -1 1",
152 "-1 -1 -1 -1",
153 "-1 -1 -1 -1",
154 "1 -1 -1 1"
155 }; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 4; verify_case(2, Arg1, getValue(Arg0)); }
156     void test_case_3() { string Arr0[] = {
157 "20 13 -2 100",
158 "-12 0 4 -3",
159 "4 1 -36 21"
160 }; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 131; verify_case(3, Arg1, getValue(Arg0)); }
161     void test_case_4() { string Arr0[] = {
162 "0"
163 }; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 0; verify_case(4, Arg1, getValue(Arg0)); }
164 
165 // END CUT HERE
166 
167 };
168 
169 // BEGIN CUT HERE
170 int main()
171 {
172 //    freopen( "a.out" , "w" , stdout );    
173     FoldThePaper ___test;
174     ___test.run_test(-1);
175        return 0;
176 }
177 // END CUT HERE
View Code
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原文地址：https://www.cnblogs.com/plumrain/p/srm_406.html