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

  1 // BEGIN CUT HERE
  2 /*
  3 
  4 */
  5 // END CUT HERE
  6 #line 7 "WinterAndPresents.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 clr0(x) memset(x, 0, sizeof(x))
 34 #define clr1(x) memset(x, -1, sizeof(x))
 35 #define pb push_back
 36 #define mp make_pair
 37 #define sz(v) ((int)(v).size())
 38 #define out(x) cout<<#x<<":"<<(x)<<endl
 39 #define tst(a) cout<<#a<<endl
 40 #define CINBEQUICKER std::ios::sync_with_stdio(false)
 41 
 42 typedef vector<int> VI;
 43 typedef vector<string> VS;
 44 typedef vector<double> VD;
 45 typedef long long int64;
 46 
 47 const double eps = 1e-8;
 48 const double PI = atan(1.0)*4;
 49 const int inf = 2139062143 / 2;
 50 
 51 inline int MyMod( int a , int b ) { return (a%b+b)%b;}
 52 
 53 int num[3000005];
 54 int num2[3000005];
 55 
 56 int64 max(int x, int64 y)
 57 {
 58     return x > y ? x : y;
 59 }
 60 
 61 class WinterAndPresents
 62 {
 63     public:
 64         long long getNumber(vector <int> an, vector <int> on){
 65             clr0 (num); clr0 (num2);
 66             int n = sz(an);
 67             for (int i = 0; i < n; ++ i){
 68                 ++ num[an[i]];
 69                 ++ num2[on[i]];
 70             }
 71             int pos = 0;
 72             for (int i = 0; i < n; ++ i)
 73                 if (an[pos] + on[pos] > an[i] + on[i]) pos = i;
 74             int64 up = an[pos] + on[pos], cnt = 1;
 75             int64 ret = 0, sum = n - num[0], maxx = 0;
 76             int64 ts = n - num2[0], minn = 0;
 77             while (cnt <= up){
 78                 maxx += sum;
 79                 minn += ts;
 80                 ret += maxx - max(0, cnt*n-minn) + 1;
 81                 sum -= num[cnt];
 82                 ts -= num2[cnt];
 83                 //out (cnt); out (ret); out (sum); out (ts);
 84                 ++ cnt;
 85             }
 86             return ret;
 87         }
 88         
 89 // BEGIN CUT HERE
 90     public:
 91     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(); }
 92     //void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0();}
 93     private:
 94     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(); }
 95     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; } }
 96     void test_case_0() { int Arr0[] = {10}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arr1[] = {0}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); long long Arg2 = 3LL; verify_case(0, Arg2, getNumber(Arg0, Arg1)); }
 97     void test_case_1() { int Arr0[] = {1, 2, 0, 3}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arr1[] = {4, 5, 0, 6}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); long long Arg2 = 0LL; verify_case(1, Arg2, getNumber(Arg0, Arg1)); }
 98     void test_case_2() { int Arr0[] = {2, 2, 2}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arr1[] = {2, 2, 2}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); long long Arg2 = 16LL; verify_case(2, Arg2, getNumber(Arg0, Arg1)); }
 99     void test_case_3() { int Arr0[] = {7, 4, 5}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arr1[] = {1, 10, 2}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); long long Arg2 = 46LL; verify_case(3, Arg2, getNumber(Arg0, Arg1)); }
100     void test_case_4() { int Arr0[] = {1000000}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arr1[] = {1000000}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); long long Arg2 = 1000002000000LL; verify_case(4, Arg2, getNumber(Arg0, Arg1)); }
101 
102 // END CUT HERE
103 
104 };
105 //by plum rain
106 // BEGIN CUT HERE
107 int main()
108 {
109     //freopen( "a.out" , "w" , stdout );    
110     WinterAndPresents ___test;
111     ___test.run_test(-1);
112        return 0;
113 }
114 // END CUT HERE
View Code

 

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

  1 // BEGIN CUT HERE
  2 /*
  3  * Author:  plum rain
  4  * score :
  5  */
  6 /*
  7 
  8  */
  9 // END CUT HERE
 10 #line 11 "WinterAndSnowmen.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 const int mod = 1000000007;
 58 
 59 int n, m;
 60 int d[2][2105][2];
 61 
 62 int gao(int r)
 63 {
 64     clr0 (d);
 65     d[0][0][0] = 1;
 66     int cur = 0, tmp = 1 << 11;
 67     for (int i = 1; i <= min(n, m); ++ i){
 68         cur ^= 1;
 69         int flag = ((i & (1<<r)) > 0);
 70         for (int j = 0; j < tmp; ++ j)
 71             for (int k = 0; k < 2; ++ k){
 72                 d[cur][j][k] = (d[cur^1][j][k] + d[cur^1][j^i][k]) % mod;
 73                 d[cur][j][k] = (d[cur][j][k] + d[cur^1][j^i][k^flag]) % mod;
 74             }
 75     }
 76     if (n > m)
 77         for (int i = m+1; i <= n; ++ i){
 78             cur ^= 1;
 79             int flag = ((i & (1<<r)) > 0);
 80             for (int j = 0; j < tmp; ++ j)
 81                 for (int k = 0; k < 2; ++ k)
 82                     d[cur][j][k] = (d[cur^1][j][k] + d[cur^1][j^i][k^flag]) % mod;
 83         }
 84     if (n < m)
 85         for (int i = n+1; i <= m; ++ i){ 
 86             cur ^= 1;
 87             for (int j = 0; j < tmp; ++ j)
 88                 for (int k = 0; k < 2; ++ k)
 89                     d[cur][j][k] = (d[cur^1][j][k] + d[cur^1][j^i][k]) % mod;
 90         }
 91     int ret = 0;
 92     for (int j = 0; j < tmp; ++ j)
 93         if ((j >> r) == 1) ret = (ret + d[cur][j][0]) % mod;
 94     return ret;
 95 }
 96 
 97 class WinterAndSnowmen
 98 {
 99     public:
100         int getNumber(int N, int M){
101             n = N; m = M;
102             int ans = 0;
103             for (int i = 0; i < 11; ++ i)
104                 ans = (ans + gao(i)) % mod;
105             return ans;
106         }
107         
108 // BEGIN CUT HERE
109     public:
110     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(); }
111     private:
112     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(); }
113     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; } }
114     void test_case_0() { int Arg0 = 2; int Arg1 = 2; int Arg2 = 4; verify_case(0, Arg2, getNumber(Arg0, Arg1)); }
115     void test_case_1() { int Arg0 = 1; int Arg1 = 1; int Arg2 = 1; verify_case(1, Arg2, getNumber(Arg0, Arg1)); }
116     void test_case_2() { int Arg0 = 3; int Arg1 = 5; int Arg2 = 74; verify_case(2, Arg2, getNumber(Arg0, Arg1)); }
117     void test_case_3() { int Arg0 = 7; int Arg1 = 4; int Arg2 = 216; verify_case(3, Arg2, getNumber(Arg0, Arg1)); }
118     void test_case_4() { int Arg0 = 47; int Arg1 = 74; int Arg2 = 962557390; verify_case(4, Arg2, getNumber(Arg0, Arg1)); }
119 
120 // END CUT HERE
121 
122 };
123 
124 // BEGIN CUT HERE
125 int main()
126 {
127 //    freopen( "a.out" , "w" , stdout );    
128     WinterAndSnowmen ___test;
129     ___test.run_test(-1);
130        return 0;
131 }
132 // END CUT HERE
View Code

 

 

posted @ 2013-12-28 23:32  Plumrain  阅读(459)  评论(0)  编辑  收藏  举报