SRM 601(1-250pt,500pt)

DIV1 250pt

题意：有很多袋子，里面装有苹果和橘子(也可能没有)，给出每个袋子里有多少个苹果，多少个橘子。如果每个袋子里含有水果的总数都不小于x个，则可以从每个袋子里都拿出x个水果(拿出苹果和橘子的总数为x)，将所有拿出的水果混合成一份礼物，问可能混合出的礼物的种数。

　　　最多50个袋子，每个袋子里的苹果和橘子的数量 <= 10^6。

解法：枚举即可，枚举从每个袋子里拿出的水果的数量x，然后求出，拿出总数是x的情况下，苹果最多能拿多少个，最少能拿多少个，相减即可。

　　　至于怎么求最多和最少，由于最多50个袋子，所以暴力即可。比赛的时候我很傻逼地用了更快的递推来求，然后赋值初始化写错了，喜大普奔。。。。。。。。。。

tag:brute-force

 // BEGIN CUT HERE
 /*
 
 */
 // END CUT HERE
 #line 7 "WinterAndPresents.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 clr0(x) memset(x, 0, sizeof(x))
 #define clr1(x) memset(x, -1, sizeof(x))
 #define pb push_back
 #define mp make_pair
 #define sz(v) ((int)(v).size())
 #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 inf =  / ;
 
 inline int MyMod( int a , int b ) { return (a%b+b)%b;}
 
 int num[];
 int num2[];
 
 int64 max(int x, int64 y)
 {
     return x > y ? x : y;
 }
 
 class WinterAndPresents
 {
     public:
         long long getNumber(vector <int> an, vector <int> on){
             clr0 (num); clr0 (num2);
             int n = sz(an);
             for (int i = ; i < n; ++ i){
                 ++ num[an[i]];
                 ++ num2[on[i]];
             }
             int pos = ;
             for (int i = ; i < n; ++ i)
                 if (an[pos] + on[pos] > an[i] + on[i]) pos = i;
             int64 up = an[pos] + on[pos], cnt = ;
             int64 ret = , sum = n - num[], maxx = ;
             int64 ts = n - num2[], minn = ;
             while (cnt <= up){
                 maxx += sum;
                 minn += ts;
                 ret += maxx - max(, cnt*n-minn) + ;
                 sum -= num[cnt];
                 ts -= num2[cnt];
                 //out (cnt); out (ret); out (sum); out (ts);
                 ++ cnt;
             }
             return 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(); if ((Case == -) || (Case == )) test_case_4(); }
     //void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0();}
     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 long long &Expected, const long long &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 Arr1[] = {}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); long long Arg2 = 3LL; verify_case(, Arg2, getNumber(Arg0, Arg1)); }
     void test_case_1() { int Arr0[] = {, , , }; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arr1[] = {, , , }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); long long Arg2 = 0LL; verify_case(, Arg2, getNumber(Arg0, Arg1)); }
     void test_case_2() { int Arr0[] = {, , }; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arr1[] = {, , }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); long long Arg2 = 16LL; verify_case(, Arg2, getNumber(Arg0, Arg1)); }
     void test_case_3() { int Arr0[] = {, , }; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arr1[] = {, , }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); long long Arg2 = 46LL; verify_case(, Arg2, getNumber(Arg0, Arg1)); }
     void test_case_4() { int Arr0[] = {}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arr1[] = {}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); long long Arg2 = 1000002000000LL; verify_case(, Arg2, getNumber(Arg0, Arg1)); }
 
 // END CUT HERE
 
 };
 //by plum rain
 // BEGIN CUT HERE
 int main()
 {
     //freopen( "a.out" , "w" , stdout );
     WinterAndPresents ___test;
     ___test.run_test(-);
        return ;
 }
 // END CUT HERE

DIV1 500pt

题意：给出整数n和m，{1,2,3....n}的一个子集a，{1,2,3...m}的一个子集b，要求a和b满足两个条件：1、a和b的元素无交集；2、a中所有元素异或得到的值小于b中所有元素异或得到的值。求这样的集合对(a, b)有多少对。

解法：首先，设a异或的值为x，b异或的值为y，由于x小于y，所以在二进制形式中，一定存在某一位r，在r位之前x和y相同，第r位x为0，y为1。枚举r。

　　　设d[i][j][k]表示前i个数，放进a或b中的所有元素的异或值为j，x第r位为k（0或1）的状态下，一共有多少个这样的集合对。然后只要取d[max(n,m)][t][0]即可，其中t满足的条件是(t >> r) == 1。注意到，本题的两个关键点，一个是用异或值为1表示，前r位x和y相同，第r位不同；另一个是，只要把一个数放进集合a或b都把它和j与一下，然后，因为在比r位高的位a和b两个集合的异或值相同，所以j的这些位异或值为0。

　　　虽然这是一个dp，但是关键的那点还是很难想到的。。。。

tag:dp, think, good

 // BEGIN CUT HERE
 /*
  * Author:  plum rain
  * score :
  */
 /*
 
  */
 // END CUT HERE
 #line 11 "WinterAndSnowmen.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 =  / ;
 const int mod = ;
 
 int n, m;
 int d[][][];
 
 int gao(int r)
 {
     clr0 (d);
     d[][][] = ;
     int cur = , tmp =  << ;
     for (int i = ; i <= min(n, m); ++ i){
         cur ^= ;
         int flag = ((i & (<<r)) > );
         for (int j = ; j < tmp; ++ j)
             for (int k = ; k < ; ++ k){
                 d[cur][j][k] = (d[cur^][j][k] + d[cur^][j^i][k]) % mod;
                 d[cur][j][k] = (d[cur][j][k] + d[cur^][j^i][k^flag]) % mod;
             }
     }
     if (n > m)
         for (int i = m+; i <= n; ++ i){
             cur ^= ;
             int flag = ((i & (<<r)) > );
             for (int j = ; j < tmp; ++ j)
                 for (int k = ; k < ; ++ k)
                     d[cur][j][k] = (d[cur^][j][k] + d[cur^][j^i][k^flag]) % mod;
         }
     if (n < m)
         for (int i = n+; i <= m; ++ i){
             cur ^= ;
             for (int j = ; j < tmp; ++ j)
                 for (int k = ; k < ; ++ k)
                     d[cur][j][k] = (d[cur^][j][k] + d[cur^][j^i][k]) % mod;
         }
     int ret = ;
     for (int j = ; j < tmp; ++ j)
         if ((j >> r) == ) ret = (ret + d[cur][j][]) % mod;
     return ret;
 }
 
 class WinterAndSnowmen
 {
     public:
         int getNumber(int N, int M){
             n = N; m = M;
             int ans = ;
             for (int i = ; i < ; ++ i)
                 ans = (ans + gao(i)) % mod;
             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 Arg0 = ; int Arg1 = ; int Arg2 = ; verify_case(, Arg2, getNumber(Arg0, Arg1)); }
     void test_case_1() { int Arg0 = ; int Arg1 = ; int Arg2 = ; verify_case(, Arg2, getNumber(Arg0, Arg1)); }
     void test_case_2() { int Arg0 = ; int Arg1 = ; int Arg2 = ; verify_case(, Arg2, getNumber(Arg0, Arg1)); }
     void test_case_3() { int Arg0 = ; int Arg1 = ; int Arg2 = ; verify_case(, Arg2, getNumber(Arg0, Arg1)); }
     void test_case_4() { int Arg0 = ; int Arg1 = ; int Arg2 = ; verify_case(, Arg2, getNumber(Arg0, Arg1)); }
 
 // END CUT HERE
 
 };
 
 // BEGIN CUT HERE
 int main()
 {
 //    freopen( "a.out" , "w" , stdout );
     WinterAndSnowmen ___test;
     ___test.run_test(-);
        return ;
 }
 // END CUT HERE