250:乱搞
解题代码：
1 // BEGIN CUT HERE 2 /* 3 4 */ 5 // END CUT HERE 6 #line 7 "RunningAroundPark.cpp" 7 #include <cstdlib> 8 #include <cctype> 9 #include <cstring> 10 #include <cstdio> 11 #include <cmath> 12 #include <algorithm> 13 #include <vector> 14 #include <string> 15 #include <iostream> 16 #include <sstream> 17 #include <map> 18 #include <set> 19 #include <queue> 20 #include <stack> 21 #include <fstream> 22 #include <numeric> 23 #include <iomanip> 24 #include <bitset> 25 #include <list> 26 #include <stdexcept> 27 #include <functional> 28 #include <utility> 29 #include <ctime> 30 using namespace std; 31 32 #define PB push_back 33 #define MP make_pair 34 35 #define REP(i,n) for(i=0;i<(n);++i) 36 #define FOR(i,l,h) for(i=(l);i<=(h);++i) 37 #define FORD(i,h,l) for(i=(h);i>=(l);--i) 38 39 typedef vector<int> VI; 40 typedef vector<string> VS; 41 typedef vector<double> VD; 42 typedef long long LL; 43 typedef pair<int,int> PII; 44 45 46 class RunningAroundPark 47 { 48 public: 49 int numberOfLap(int N, vector <int> d) 50 { 51 int hs[100]; 52 memset(hs,0,sizeof(hs)); 53 int len = d.size(); 54 int sum = 1; 55 for(int i = 1 ;i < len ;i ++) 56 { 57 if(d[i] <= d[i-1] ) 58 sum ++ ; 59 } 60 return sum ; 61 } 62 63 64 };
500：给定序列的二进制尾巴0的个数，问你这一序列的子序列可能构成等比数列的可能数。
解题思路：子序列为等比数列的唯一可能性是这个序列二进制后缀0个数为等差数列
解题代码：
1 // BEGIN CUT HERE 2 /* 3 4 */ 5 // END CUT HERE 6 #line 7 "PotentialGeometricSequence.cpp" 7 #include <cstdlib> 8 #include <cctype> 9 #include <cstring> 10 #include <cstdio> 11 #include <cmath> 12 #include <algorithm> 13 #include <vector> 14 #include <string> 15 #include <iostream> 16 #include <sstream> 17 #include <map> 18 #include <set> 19 #include <queue> 20 #include <stack> 21 #include <fstream> 22 #include <numeric> 23 #include <iomanip> 24 #include <bitset> 25 #include <list> 26 #include <stdexcept> 27 #include <functional> 28 #include <utility> 29 #include <ctime> 30 using namespace std; 31 32 #define PB push_back 33 #define MP make_pair 34 35 #define REP(i,n) for(i=0;i<(n);++i) 36 #define FOR(i,l,h) for(i=(l);i<=(h);++i) 37 #define FORD(i,h,l) for(i=(h);i>=(l);--i) 38 39 typedef vector<int> VI; 40 typedef vector<string> VS; 41 typedef vector<double> VD; 42 typedef long long LL; 43 typedef pair<int,int> PII; 44 45 46 class PotentialGeometricSequence 47 { 48 public: 49 int numberOfSubsequences(vector <int> d) 50 { 51 int len = d.size(); 52 int sum = len; 53 for(int i = 1 ;i < len ;i ++) 54 { 55 int temp = d[i] - d[i-1]; 56 for(int j = i ;j < len ;j ++) 57 { 58 int ok = 1 ; 59 for(int s = i ;s <= j; s ++) 60 { 61 if(d[s] - d[s-1] != temp) 62 ok = 0 ; 63 } 64 if(ok) 65 sum ++ ; 66 } 67 } 68 return sum; 69 } 70 71 72 };
1000：问你100个数选任意多数乘积为K 的总数有多少
解题思路：可以知道 k的约数最多为1000个 ，我们用map背包就行
解题代码：
1 // BEGIN CUT HERE 2 /* 3 4 */ 5 // END CUT HERE 6 #line 7 "GoodSubset.cpp" 7 #include <cstdlib> 8 #include <cctype> 9 #include <cstring> 10 #include <cstdio> 11 #include <cmath> 12 #include <algorithm> 13 #include <vector> 14 #include <string> 15 #include <iostream> 16 #include <sstream> 17 #include <map> 18 #include <set> 19 #include <queue> 20 #include <stack> 21 #include <fstream> 22 #include <numeric> 23 #include <iomanip> 24 #include <bitset> 25 #include <list> 26 #include <stdexcept> 27 #include <functional> 28 #include <utility> 29 #include <ctime> 30 using namespace std; 31 32 #define PB push_back 33 #define MP make_pair 34 35 #define REP(i,n) for(i=0;i<(n);++i) 36 #define FOR(i,l,h) for(i=(l);i<=(h);++i) 37 #define FORD(i,h,l) for(i=(h);i>=(l);--i) 38 39 typedef vector<int> VI; 40 typedef vector<string> VS; 41 typedef vector<double> VD; 42 typedef long long LL; 43 typedef pair<int,int> PII; 44 45 vector <LL>a; 46 int num[105]; 47 int n ; 48 int rn; 49 map <LL,LL> mp[105]; 50 map <LL,LL> ::iterator it; 51 #define M 1000000007 52 class GoodSubset 53 { 54 public: 55 int numberOfSubsets(int goodValue, vector <int> d) 56 { 57 for(int i = 0 ;i <=100 ;i ++) 58 mp[i].clear(); 59 mp[0][1] = 1; 60 n = d.size(); 61 for(int i = 1 ;i <=n ;i ++) 62 { 63 int t = 0 ; 64 for( it = mp[i-1].begin();it != mp[i-1].end();it++) 65 { 66 t ++ ; 67 //printf("%d %lld %lld ",i,it->first,it->second); 68 mp[i][it->first] = (mp[i][it->first] + mp[i-1][it->first])%M; 69 if(goodValue % (d[i-1]*it->first) == 0 ) 70 mp[i][d[i-1]*it->first] = (it->second+mp[i][d[i-1]*it->first])%M; 71 } 72 // printf("%d ",t); 73 } 74 if(goodValue == 1) 75 mp[n][goodValue] -- ; 76 return int(mp[n][goodValue]); 77 } 78 79 };