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
// BEGIN CUT HERE
/*
* Author: plum rain
* score :
*/
/* */
// END CUT HERE
#line 11 "SymmetricPie.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 clr0(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<<" "
#define tst1(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 inf = / ; int n, ans;
VI an, dn;
bool mp[];
int sum[]; void dfs(int x)
{
if (x == n){
clr0 (sum); sum[] = dn[an[]];
for (int i = ; i < n; ++ i)
sum[i] = sum[i-] + dn[an[i]]; int cnt = ;
for (int i = ; i < n; ++ i)
for (int j = ; j < n; ++ j){
int tmp = j > i ? sum[j] - sum[i] : sum[n-] - sum[i] + sum[j];
if (tmp == ) ++ cnt;
}
ans = max(ans, cnt/);
return ;
} for (int i = ; i < n; ++ i) if (!mp[i]){
an.pb (i); mp[i] = ;
dfs (x + );
an.pop_back(); mp[i] = ;
}
} class SymmetricPie
{
public:
int getLines(vector <int> D){
dn = D;
ans = ; n = sz(dn);
clr0 (mp); an.clear();
dfs ();
return ans;
} // 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(); if ((Case == -) || (Case == )) test_case_4(); }
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 = ; verify_case(, Arg1, getLines(Arg0)); }
void test_case_1() { int Arr0[] = {,,}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; verify_case(, Arg1, getLines(Arg0)); }
void test_case_2() { int Arr0[] = {,}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; verify_case(, Arg1, getLines(Arg0)); }
void test_case_3() { int Arr0[] = {,,,,,}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; verify_case(, Arg1, getLines(Arg0)); }
void test_case_4() { int Arr0[] = {,,}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; verify_case(, Arg1, getLines(Arg0)); } // END CUT HERE }; // BEGIN CUT HERE
int main()
{
// freopen( "a.out" , "w" , stdout );
SymmetricPie ___test;
___test.run_test(-);
return ;
}
// END CUT HERE
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
// BEGIN CUT HERE
/*
* Author: plum rain
* score :
*/
/* */
// END CUT HERE
#line 11 "FoldThePaper.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 clr0(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<<" "
//#define tst1(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 inf = / ; int n, m;
int an[][];
bool can[<<];
vector<vi> row, col; vi getvec(int x, int f)
{
vi v;
for (int i = ; i < (f ? m : n); ++ i)
if (x & (<<i)) v.pb (i);
return v;
} void rec(vi v){
//for (int i = 0; i < sz(v); ++ i)
//tst (v[i]);
//cout << endl;
int n = sz(v);
for (int i = ; i < n; ++ i)
can[v[i]] = ;
for (int i = ; i < n; ++ i){
vi nv(max(i, n-i));
if (i <= n/){
for (int j = i; j < n; ++ j)
nv[j-i] = v[j];
for (int j = , k = i-; j < i; ++ j, -- k)
nv[j] |= v[k];
}
else{
for (int j = ; j < i; ++ j)
nv[j] = v[j];
for (int j = i, k = i-; j < n; ++ j, -- k)
nv[k] |= v[j];
}
rec(nv);
}
} class FoldThePaper
{
public:
int getValue(vector <string> pap){
n = sz(pap);
stringstream stm;
for (int i = ; i < n; ++ i){
stm << pap[i]; m = ;
while (stm >> an[i][m]) ++ m;
stm.clear();
}
clr0 (can);
vi v;
for (int i = ; i < n; ++ i) v.pb (<<i);
rec(v);
row.clear();
for (int i = ; i < (<<n); ++ i) if (can[i])
row.pb (getvec(i, )); v.clear(); clr0 (can);
for (int i = ; i < m; ++ i) v.pb (<<i);
rec(v);
col.clear();
for (int i = ; i < (<<m); ++ i) if (can[i])
col.pb (getvec(i, )); int ans = -inf;
for (int i = ; i < sz(row); ++ i)
for (int j = ; j < sz(col); ++ j){
int tmp = -inf;
for (int t = ; t < sz(row[i]); ++ t)
for (int k = ; k < sz(col[j]); ++ k){
if (tmp == -inf) tmp = an[row[i][t]][col[j][k]];
else tmp += an[row[i][t]][col[j][k]];
}
ans = max(tmp, ans);
}
return ans;
} // 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(); if ((Case == -) || (Case == )) test_case_4(); }
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() { string Arr0[] = {
"1 1 1",
"1 1 1"
}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; verify_case(, Arg1, getValue(Arg0)); }
void test_case_1() { string Arr0[] = {
"1 -1",
"1 -1"
}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; verify_case(, Arg1, getValue(Arg0)); }
void test_case_2() { string Arr0[] = {
"1 -1 -1 1",
"-1 -1 -1 -1",
"-1 -1 -1 -1",
"1 -1 -1 1"
}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; verify_case(, Arg1, getValue(Arg0)); }
void test_case_3() { string Arr0[] = {
"20 13 -2 100",
"-12 0 4 -3",
"4 1 -36 21"
}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; verify_case(, Arg1, getValue(Arg0)); }
void test_case_4() { string Arr0[] = {
""
}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; verify_case(, Arg1, getValue(Arg0)); } // END CUT HERE }; // BEGIN CUT HERE
int main()
{
// freopen( "a.out" , "w" , stdout );
FoldThePaper ___test;
___test.run_test(-);
return ;
}
// END CUT HERE