水题合集

牛客小白月赛12 392 

E  签到题  二分答案

#include <iostream>
#include <cstring>
#include <cstdio>
#include <string>
#include <queue>
#include <list>
#include <map>
#include <set>
#include <cmath>
#include <bitset>
#include <vector>
#include <iomanip>
#include <sstream>
#include <cstdlib>
#include <algorithm>
using namespace std;
typedef long long  ll;
#define mem(A, X) memset(A, X, sizeof A)
#define foreach(e,x) for(__typeof(x.begin()) e=x.begin();e!=x.end();++e)
#define fori(i,l,u) for(ll (i)=(ll)(l);(i)<=(ll)(u);++(i))
#define ford(i,l,u) for(ll (i)=(ll)(l);(i)>=(ll)(u);--(i))

ll n,k;
int a[200005];

bool check(int len){
    ll cnt=0;
    fori(i,1,n) cnt+=a[i]/len;
    return cnt>=k;
}
void solve(){

    int l=0,r=1e9+10;
    //符和（某种情况）的最大值
    while(l<r)
    {
        int m=(l+r+1)/2;
        if(check(m))
            l=m;
        else
            r=m-1;
    }
    cout<<l<<endl;

}
int main()
{
  ios::sync_with_stdio(false);
  //freopen("local.in","r",stdin);
  while(cin>>n>>k){
      fori(i,1,n) cin>>a[i];
      solve();
  }



return 0;
}

 

G  可以证明，答案是gcd（n,m）

#include <iostream>
#include <cstring>
#include <cstdio>
#include <string>
#include <queue>
#include <list>
#include <map>
#include <set>
#include <cmath>
#include <bitset>
#include <vector>
#include <iomanip>
#include <sstream>
#include <cstdlib>
#include <algorithm>
using namespace std;
typedef long long  ll;
#define mem(A, X) memset(A, X, sizeof A)
#define foreach(e,x) for(__typeof(x.begin()) e=x.begin();e!=x.end();++e)
#define fori(i,l,u) for(ll (i)=(ll)(l);(i)<=(ll)(u);++(i))
#define ford(i,l,u) for(ll (i)=(ll)(l);(i)>=(ll)(u);--(i))

ll gcd(ll x, ll y){ return y? gcd(y,x%y) : x;}
ll a,b;
string s;
int main()
{
  ios::sync_with_stdio(false);
  //freopen("local.in","r",stdin);
  while(cin>>a>>b>>s){
      cout<<gcd(a,b)<<endl;
  }

return 0;
}