1、水题，直接穷举就OK了。
1 #include <iostream> 2 #include <string> 3 #include <vector> 4 #include <cstdlib> 5 #include <cmath> 6 #include <map> 7 #include <stack> 8 #include <algorithm> 9 #include <list> 10 #include <ctime> 11 #include <set> 12 #include <queue> 13 #define kfor(i,s,e) for(int i=s;i<e;i++) 14 typedef long long ll; 15 using namespace std; 16 17 class TheJediTestDiv2 { 18 public: 19 int countSupervisors(vector<int> students, int Y, int J) { 20 int ct = 0; 21 int res = 5000000; 22 vector<int> studentscopy = students; 23 for (int i = 0; i < students.size(); i++) { 24 students[i] = ((max(0, students[i] - Y)) + (J - 1)) / J; 25 studentscopy[i] = ((max(0, students[i])) + (J - 1)) / J; 26 ct = ct + students[i]; 27 cout<<ct<<endl; 28 } 29 for (int i = 0; i < students.size(); i++) { 30 res = min(res, ct - studentscopy[i] + students[i]); 31 } 32 return res; 33 } 34 };
2、找规律即可，若想成功区分，最少是110。
1 #include <iostream> 2 #include <string> 3 #include <vector> 4 #include <cstdlib> 5 #include <cmath> 6 #include <map> 7 #include <stack> 8 #include <algorithm> 9 #include <list> 10 #include <ctime> 11 #include <set> 12 #include <queue> 13 #define kfor(i,s,e) for(int i=s;i<e;i++) 14 typedef long long ll; 15 using namespace std; 16 17 class TheDeviceDiv2 { 18 public: 19 string identify(vector <string> plates) { 20 int r=plates.size(); 21 int c=plates[0].size(); 22 for(int i=0;i<c;i++){ 23 int one,zero; 24 one=0;zero=0; 25 for(int j=0;j<r;j++){ 26 if(plates[j][i]=='1'){ 27 one++; 28 }else{ 29 zero++; 30 } 31 } 32 if(one<2||zero<1){ 33 return "NO"; 34 } 35 } 36 return "YES"; 37 } 38 };
3、DP的问题，看的wiki上的答案。
1 #include <iostream> 2 #include <string> 3 #include <vector> 4 #include <cstdlib> 5 #include <cmath> 6 #include <map> 7 #include <stack> 8 #include <algorithm> 9 #include <list> 10 #include <ctime> 11 #include <set> 12 #include <queue> 13 typedef long long ll; 14 using namespace std; 15 ll dp[1205][1205]; 16 bool composite[1105]; 17 ll MOD = 1000000009; 18 19 20 class MegaFactorialDiv2 { 21 public: 22 ll primeAppears(int N, int K, int p) 23 { 24 ll res = 0; 25 // For each factor x: 26 for (int x = p ; x <= N; x+=p) { 27 int c = 0; 28 int y = x; 29 while ( y%p == 0) { 30 y /= p; 31 c++; 32 } 33 if (c > 0) { 34 //how many times does this factor appear? 35 res += (c * dp[K + N - x - 1][N - x] ) % MOD; 36 } 37 } 38 return res % MOD; 39 } 40 41 42 int countDivisors(int N, int K) 43 { 44 // Precalculate binomial coefficients using Pascal's triangle 45 for (int i=0; i<=K + N; i++) { 46 dp[i][0] = 1; 47 for (int j=1; j<=i; j++) { 48 dp[i][j] = ( dp[i-1][j-1] + dp[i-1][j] ) % MOD; 49 } 50 } 51 52 //for each prime between 1 and N: How many times does it appear in N!K? 53 ll res = 1; 54 // For each prime number i between 1 and N: (Use Erathostenes' Sieve) 55 for (int i=2; i<=N; i++) { 56 if (! composite[i]) { 57 for (int j=i+i; j<=N; j+=i) { 58 composite[j] = true; 59 } 60 res = (res * (primeAppears(N,K, i ) + 1) ) % MOD; 61 } 62 } 63 return (int)( res % MOD ); 64 65 } 66 };