srm 569 div2

1、水题，直接穷举就OK了。

#include <iostream>
#include <string>
#include <vector>
#include <cstdlib>
#include <cmath>
#include <map>
#include <stack>
#include <algorithm>
#include <list>
#include <ctime>
#include <set>
#include <queue>
#define kfor(i,s,e)  for(int i=s;i<e;i++)
typedef long long ll;
using namespace std;

class TheJediTestDiv2 {
public:
    int countSupervisors(vector<int> students, int Y, int J) {
        int ct = 0;
        int res = 5000000;
        vector<int> studentscopy = students;
        for (int i = 0; i < students.size(); i++) {
            students[i] = ((max(0, students[i] - Y)) + (J - 1)) / J;
            studentscopy[i] = ((max(0, students[i])) + (J - 1)) / J;
            ct = ct + students[i];
            cout<<ct<<endl;
        }
        for (int i = 0; i < students.size(); i++) {
            res = min(res, ct - studentscopy[i] + students[i]);
        }
        return res;
    }
};

2、找规律即可，若想成功区分，最少是110。

#include <iostream>
#include <string>
#include <vector>
#include <cstdlib>
#include <cmath>
#include <map>
#include <stack>
#include <algorithm>
#include <list>
#include <ctime>
#include <set>
#include <queue>
#define kfor(i,s,e)  for(int i=s;i<e;i++)
typedef long long ll;
using namespace std;

class TheDeviceDiv2 {
public:
    string identify(vector <string> plates) {
        int r=plates.size();
        int c=plates[0].size();
        for(int i=0;i<c;i++){
            int one,zero;
            one=0;zero=0;
            for(int j=0;j<r;j++){
                if(plates[j][i]=='1'){
                    one++;
                }else{
                    zero++;
                }
            }
            if(one<2||zero<1){
                return "NO";
            }
        }
        return "YES";
    }
};

3、DP的问题，看的wiki上的答案。

#include <iostream>
#include <string>
#include <vector>
#include <cstdlib>
#include <cmath>
#include <map>
#include <stack>
#include <algorithm>
#include <list>
#include <ctime>
#include <set>
#include <queue>
typedef long long ll;
using namespace std;
ll dp[1205][1205];
bool composite[1105];
ll MOD = 1000000009;


class MegaFactorialDiv2 {
public:
    ll primeAppears(int N, int K, int p)
    {
        ll res = 0;
        // For each factor x:
        for (int x = p ; x <= N; x+=p) {
            int c = 0;
            int y = x;
            while ( y%p == 0) {
                y /= p;
                c++;
            }
            if (c > 0) {
                //how many times does this factor appear?
                res += (c * dp[K + N - x - 1][N - x]  ) % MOD;
            }
        }
        return res % MOD;
    }


    int countDivisors(int N, int K)
    {
        // Precalculate binomial coefficients using Pascal's triangle
        for (int i=0; i<=K + N; i++) {
            dp[i][0] = 1;
            for (int j=1; j<=i; j++) {
                dp[i][j] = ( dp[i-1][j-1] + dp[i-1][j] ) % MOD;
            }
        }

        //for each prime between 1 and N: How many times does it appear in N!K?
        ll res = 1;
        // For each prime number i between 1 and N: (Use Erathostenes' Sieve)
        for (int i=2; i<=N; i++) {
            if (! composite[i]) {
                for (int j=i+i; j<=N; j+=i) {
                    composite[j] = true;
                }
                res  = (res * (primeAppears(N,K, i ) + 1) ) % MOD;
            }
        }
        return (int)( res % MOD );

    }
};