久违的cf服务器爆炸场

A. Odd Divisor

思路：任何一个数都可以写成\(n = k2^m，其中k是一个奇数\)，若k=1，那么n就一定是一个2的幂。

view code 
#include<iostream>

#include<string>

#include<algorithm>

#include<cstdio>

#include<cstring>

#include<cmath>

#include<map>

#include <queue>

#include<sstream>

#include <stack>

#include <set>

#include <bitset>

#include<vector>

#define FAST ios::sync_with_stdio(false)

#define abs(a) ((a)>=0?(a):-(a))

#define sz(x) ((int)(x).size())

#define all(x) (x).begin(),(x).end()

#define mem(a,b) memset(a,b,sizeof(a))

#define max(a,b) ((a)>(b)?(a):(b))

#define min(a,b) ((a)<(b)?(a):(b))

#define rep(i,a,n) for(int i=a;i<=n;++i)

#define per(i,n,a) for(int i=n;i>=a;--i)

#define endl '\n'

#define pb push_back

#define mp make_pair

#define fi first

#define se second

using namespace std;

typedef long long ll;

typedef pair<ll,ll> PII;

const int maxn = 1e5+200;

const int inf=0x3f3f3f3f;

const double eps = 1e-7;

const double pi=acos(-1.0);

const int mod = 1e9+7;

inline int lowbit(int x){return x&(-x);}

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);

inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}

inline ll inv(ll x,ll p){return qpow(x,p-2,p);}

inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}

inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0',  ch = getchar();return x*f; }

int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };

int main()

{

    int kase;

    cin>>kase;

    while(kase--)

    {

        ll n = read();

        int flag = 1;

        for(ll i=1; (1LL<<i)<=n; i++ )

        {

            if((1LL<<i)==n) flag = 0;

        }

        puts(flag?"YES":"NO");

    }

    return 0;

}

B. New Year's Number

思路：你知道答案肯定是\(n = 2020i + 2021j\)的形式，那么你只需要枚举i，看j是否存在即可。

view code 
#include<iostream>

#include<string>

#include<algorithm>

#include<cstdio>

#include<cstring>

#include<cmath>

#include<map>

#include <queue>

#include<sstream>

#include <stack>

#include <set>

#include <bitset>

#include<vector>

#define FAST ios::sync_with_stdio(false)

#define abs(a) ((a)>=0?(a):-(a))

#define sz(x) ((int)(x).size())

#define all(x) (x).begin(),(x).end()

#define mem(a,b) memset(a,b,sizeof(a))

#define max(a,b) ((a)>(b)?(a):(b))

#define min(a,b) ((a)<(b)?(a):(b))

#define rep(i,a,n) for(int i=a;i<=n;++i)

#define per(i,n,a) for(int i=n;i>=a;--i)

#define endl '\n'

#define pb push_back

#define mp make_pair

#define fi first

#define se second

using namespace std;

typedef long long ll;

typedef pair<ll,ll> PII;

const int maxn = 1e5+200;

const int inf=0x3f3f3f3f;

const double eps = 1e-7;

const double pi=acos(-1.0);

const int mod = 1e9+7;

inline int lowbit(int x){return x&(-x);}

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);

inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}

inline ll inv(ll x,ll p){return qpow(x,p-2,p);}

inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}

inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0',  ch = getchar();return x*f; }

int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };

int main()

{

    int kase;

    cin>>kase;

    while(kase--)

    {

        ll n = read();

        int flag = 0;

        for(ll i=0; i*2020<=n; i++)

        {

            ll cur = n - i*2020;

            double num = cur/2021.0;

            if(num==(ll)num)

            {

                flag = 1;

                break;

            }

        }

        puts(flag?"YES":"NO");

    }

    return 0;

}

C. Ball in Berland

思路：因为只选出两组，我们只需要枚举其中一组的选法，另外一组就自然确定了。枚举a中每个人的舞伴时，当前贡献就是k - 和当前这个a有的舞伴个数 - 要和a跳舞的那个人的舞伴个数 + 本次的1， 画个图自己理解一下就很容易懂了。 最后答案要除2因为会有重复计算的对。

view code 
#include<iostream>

#include<string>

#include<algorithm>

#include<cstdio>

#include<cstring>

#include<cmath>

#include<map>

#include <queue>

#include<sstream>

#include <stack>

#include <set>

#include <bitset>

#include<vector>

#define FAST ios::sync_with_stdio(false)

#define abs(a) ((a)>=0?(a):-(a))

#define sz(x) ((int)(x).size())

#define all(x) (x).begin(),(x).end()

#define mem(a,b) memset(a,b,sizeof(a))

#define max(a,b) ((a)>(b)?(a):(b))

#define min(a,b) ((a)<(b)?(a):(b))

#define rep(i,a,n) for(int i=a;i<=n;++i)

#define per(i,n,a) for(int i=n;i>=a;--i)

#define endl '\n'

#define pb push_back

#define mp make_pair

#define fi first

#define se second

using namespace std;

typedef long long ll;

typedef pair<ll,ll> PII;

const int maxn = 2e5+200;

const int inf=0x3f3f3f3f;

const double eps = 1e-7;

const double pi=acos(-1.0);

const int mod = 1e9+7;

inline int lowbit(int x){return x&(-x);}

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);

inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}

inline ll inv(ll x,ll p){return qpow(x,p-2,p);}

inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}

inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0',  ch = getchar();return x*f; }

int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };

vector<vector<ll> > a(maxn);

vector<vector<ll> > b(maxn);

ll ina[maxn];

ll inb[maxn];

int main()

{

    int kase;

    cin>>kase;

    int num = 1;

    while(kase--)

    {

        ll na = read(), nb = read(), k = read();

        for(int i=0; i<=na+2; i++)

        a[i].clear();

        for(int i=0; i<=nb+2; i++)

        b[i].clear();

        rep(i,1,k)  ina[i] = read();

        rep(i,1,k)  inb[i] = read();

        rep(i,1,k)

        {

            a[ina[i]].pb(inb[i]);

            b[inb[i]].pb(ina[i]);

        }

        ll ans = 0;

        rep(i,1,na)

        {

            for(int j=0; j<a[i].size(); j++)

            {

                ll cur = a[i][j];

                ans += k-a[i].size()-b[cur].size()+1;

            }

        }

        cout<<ans/2<<endl;

        num++;

    }

    return 0;

}

D. Cleaning the Phone

思路：破题口在于b只有两种情况，\(bi=1或bi=2\)，那么我们可以采取前面B题的思路，答案肯定是选了x个bi=1的 +　ｙ个bi=2的， 由于贪心策略，我在这两种b的选择中，如果选k个，那肯定选a最大的那k个。 所以就只需要从大到小排序好，记录各自前缀和后，枚举一遍x，用二分看还需要多少个y，然后每次比较取最小即可。

view code 
    #include<iostream>

    #include<string>

    #include<algorithm>

    #include<cstdio>

    #include<cstring>

    #include<cmath>

    #include<map>

    #include <queue>

    #include<sstream>

    #include <stack>

    #include <set>

    #include <bitset>

    #include<vector>

    #define FAST ios::sync_with_stdio(false)

    #define abs(a) ((a)>=0?(a):-(a))

    #define sz(x) ((int)(x).size())

    #define all(x) (x).begin(),(x).end()

    #define mem(a,b) memset(a,b,sizeof(a))

    #define max(a,b) ((a)>(b)?(a):(b))

    #define min(a,b) ((a)<(b)?(a):(b))

    #define rep(i,a,n) for(int i=a;i<=n;++i)

    #define per(i,n,a) for(int i=n;i>=a;--i)

    #define endl '\n'

    #define pb push_back

    #define mp make_pair

    #define fi first

    #define se second

    using namespace std;

    typedef long long ll;

    typedef pair<ll,ll> PII;

    const int maxn = 2e5+200;

    const int inf=0x3f3f3f3f;

    const double eps = 1e-7;

    const double pi=acos(-1.0);

    const int mod = 1e9+7;

    inline int lowbit(int x){return x&(-x);}

    ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

    void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);

    inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}

    inline ll inv(ll x,ll p){return qpow(x,p-2,p);}

    inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}

    inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0',  ch = getchar();return x*f; }

    int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };

    ll a[maxn];

    ll b[maxn];

    ll suma[maxn];

    ll sumb[maxn];

    int main()

    {

        int kase;

        cin>>kase;

        while(kase--)

        {

            ll n = read(), m = read();

            rep(i,1,n) a[i] = read();

            rep(i,1,n) b[i] = read();

            vector<ll> one;

            vector<ll> two;

            rep(i,1,n)

            {

                if(b[i]==1) one.pb(a[i]);

                else two.pb(a[i]);

            }

            sort(one.begin(),one.end(), greater<ll>());

            sort(two.begin(),two.end(),greater<ll>());

            int lena = one.size();

            int lenb = two.size();

            rep(i,1,lena) suma[i] = suma[i-1] + one[i-1];

            rep(i,1,lenb) sumb[i] = sumb[i-1] + two[i-1];

            ll ans = 1e18;

            rep(i,0,lena)

            {

                ll One = suma[i];

                ll need = m - suma[i];

                if(need<=0)

                {

                    ans = min(ans,i);

                }

                else

                {

                    int id = lower_bound(sumb+1,sumb+1+lenb, need) - sumb;

                    if(id>lenb) continue;

                    ans = min(ans,id*2+i);

                }

            }

            if(ans==1e18) cout<<-1<<endl;

            else cout<<ans<<endl;

        }

        return 0;

    }

E. Advertising Agency

水题，预处理一下组合数就行了。我肯定挑的是最大的那k个，每次的贡献就是ans = ans*C(a[i]的个数，要挑出的个数)。

view code 
#include<iostream>

#include<string>

#include<algorithm>

#include<cstdio>

#include<cstring>

#include<cmath>

#include<map>

#include <queue>

#include<sstream>

#include <stack>

#include <set>

#include <bitset>

#include<vector>

#define FAST ios::sync_with_stdio(false)

#define abs(a) ((a)>=0?(a):-(a))

#define sz(x) ((int)(x).size())

#define all(x) (x).begin(),(x).end()

#define mem(a,b) memset(a,b,sizeof(a))

#define max(a,b) ((a)>(b)?(a):(b))

#define min(a,b) ((a)<(b)?(a):(b))

#define rep(i,a,n) for(int i=a;i<=n;++i)

#define per(i,n,a) for(int i=n;i>=a;--i)

#define endl '\n'

#define pb push_back

#define mp make_pair

#define fi first

#define se second

using namespace std;

typedef long long ll;

typedef pair<ll,ll> PII;

const int maxn = 1e5+200;

const int inf=0x3f3f3f3f;

const double eps = 1e-7;

const double pi=acos(-1.0);

const int mod = 1e9+7;

inline int lowbit(int x){return x&(-x);}

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);

inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}

inline ll inv(ll x,ll p){return qpow(x,p-2,p);}

inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}

inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0',  ch = getchar();return x*f; }

int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };

ll dp[1005][1005];

ll a[maxn];

void init()

{

    rep(i,0,1003) dp[i][0] = 1;

    rep(i,1,1003) rep(j,1,1003) dp[i][j] = (dp[i-1][j-1] + dp[i-1][j])%mod;

}

int main()

{

    init();

    int kase;

    cin>>kase;

    while(kase--)

    {

        map<ll,ll> Map;

        ll n = read(), k =  read();

        rep(i,1,n) a[i] = read(), Map[a[i]]++;

        sort(a+1,a+1+n);

        ll pre = -1;

        ll ans = 1;

        per(i,n,1)

        {

            if(a[i]==pre) continue;

            ll d = min(k,Map[a[i]]);

            ans = ((ans%mod)*(dp[Map[a[i]]][d]%mod))%mod;

            k -= d;

            pre = a[i];

        }

        cout<<ans<<endl;

    }

    return 0;

}

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