Educational Codeforces Round 86 (Rated for Div. 2)部分题解

Problem A

不开ll见祖宗

Problem B

要求长度小于等于2n那么，我们贪心的情况下肯定选择最多的那个，然后直接补，这个长度一定不会超过2n，所以直接输出就好了，循环k值在不是一的情况下一定可以被补成2.

#include<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
#define rep(i,f_start,f_end) for (int i=f_start;i<=f_end;++i)
#define per(i,n,a) for (int i=n;i>=a;i--)
#define MT(x,i) memset(x,i,sizeof(x) )
#define rev(i,start,end) for (int i=start;i<end;i++)
#define inf 0x3f3f3f3f
#define mp(x,y) make_pair(x,y)
#define lowbit(x) (x&-x)
#define MOD 1000000007
#define exp 1e-8
#define N 1000005 
#define fi first 
#define se second
#define pb push_back
typedef long long ll;
const ll INF=0x3f3f3f3f3f3f3f3f;
typedef vector <int> VI;
typedef pair<int ,int> PII;
typedef pair<int ,PII> PIII;
ll gcd(ll a,ll b){return b==0?a:gcd(b,a%b);}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;};
void check_max (int &a,int b) { a=max (a,b);}
void check_min (int &a,int b) { a=min (a,b);}
inline int read() {
    char ch=getchar(); int x=0, f=1;
    while(ch<'0'||ch>'9') {
        if(ch=='-') f=-1;
        ch=getchar();
    } while('0'<=ch&&ch<='9') {
        x=x*10+ch-'0';
        ch=getchar();
    } return x*f;
}
const int maxn=1e3+10;
int t;
string str;

void solve () {
    cin>>str;
    map <int,int> mp;
    for (auto it:str) mp[it-'0']++;
    if (!mp[0]||!mp[1]||str.size ()==2)  cout<<str<<endl;
    else {
        cout<<str[0];
        rev (i,1,str.size ()) {
            if (str[i]==str[i-1]) {
                if (str[i]=='1') cout<<"0";
                else cout<<"1";
            }
            cout<<str[i];
        }
        cout<<endl;
    } 
}

int main () {
    ios::sync_with_stdio (false);
    cin>>t;
    while (t--) {
        solve ();
    }
    return 0;
}

Problem C

出现了区间个数，那么c题的位置应该就是类似前缀和的操作，那么我们很容易可以猜到推出，这些个数是以ab为一个周期的，那么我们只需要对0-a b做一个完全剩余系的前缀和统计就好了。

#include<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
#define rep(i,f_start,f_end) for (int i=f_start;i<=f_end;++i)
#define per(i,n,a) for (int i=n;i>=a;i--)
#define MT(x,i) memset(x,i,sizeof(x) )
#define rev(i,start,end) for (int i=start;i<end;i++)
#define inf 0x3f3f3f3f
#define mp(x,y) make_pair(x,y)
#define lowbit(x) (x&-x)
#define MOD 1000000007
#define exp 1e-8
#define N 1000005 
#define fi first 
#define se second
#define pb push_back
typedef long long ll;
const ll INF=0x3f3f3f3f3f3f3f3f;
typedef vector <int> VI;
typedef pair<int ,int> PII;
typedef pair<int ,PII> PIII;
ll gcd(ll a,ll b){return b==0?a:gcd(b,a%b);}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;};
void check_max (int &a,int b) { a=max (a,b);}
void check_min (int &a,int b) { a=min (a,b);}
inline int read() {
    char ch=getchar(); int x=0, f=1;
    while(ch<'0'||ch>'9') {
        if(ch=='-') f=-1;
        ch=getchar();
    } while('0'<=ch&&ch<='9') {
        x=x*10+ch-'0';
        ch=getchar();
    } return x*f;
}
const int maxn=1e3+10;
int t,a,b,q;
vector <ll> pre;

ll calc (ll x) {
    return x/(a*b)*pre[a*b]+pre[x%(a*b)];
}

void solve () {
   scanf ("%d%d%d",&a,&b,&q);
   pre.resize (a*b+1,0);
   rev (i,0,a*b) pre[i+1]=pre[i]+(i%a%b!=i%b%a);
   while (q--) {
       ll l,r;
       scanf ("%lld%lld",&l,&r);
       printf ("%lld ",calc (r+1)-calc (l));
   }
   printf ("
");
}

int main () {
    scanf ("%d",&t);
    while (t--) {
        solve ();
    }
    return 0;
}

Problem D

这个题目，是类似一个模拟加贪心的思路，我们从后往前遍历，当c值小于前一个值的时候就去重新从0号集合开始放（c值允许放，重点是给出的c数组单调下降），因为从大的开始放，先开始哪个必然是最大的，以后放的话c值都会增大，然后模拟一个二维数组当作集合就好了。

#include<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
#define rep(i,f_start,f_end) for (int i=f_start;i<=f_end;++i)
#define per(i,n,a) for (int i=n;i>=a;i--)
#define MT(x,i) memset(x,i,sizeof(x) )
#define rev(i,start,end) for (int i=start;i<end;i++)
#define inf 0x3f3f3f3f
#define mp(x,y) make_pair(x,y)
#define lowbit(x) (x&-x)
#define MOD 1000000007
#define exp 1e-8
#define N 1000005 
#define fi first 
#define se second
#define pb push_back
typedef long long ll;
const ll INF=0x3f3f3f3f3f3f3f3f;
typedef vector <int> VI;
typedef pair<int ,int> PII;
typedef pair<int ,PII> PIII;
ll gcd(ll a,ll b){return b==0?a:gcd(b,a%b);}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;};
void check_max (int &a,int b) { a=max (a,b);}
void check_min (int &a,int b) { a=min (a,b);}
inline int read() {
    char ch=getchar(); int x=0, f=1;
    while(ch<'0'||ch>'9') {
        if(ch=='-') f=-1;
        ch=getchar();
    } while('0'<=ch&&ch<='9') {
        x=x*10+ch-'0';
        ch=getchar();
    } return x*f;
}
const int maxn=1e3+10;
int t;

void solve () {
    int n,k;
    scanf ("%d%d",&n,&k);
    vector <int> cnt (k);
    rep (i,1,n) {
        int x;
        scanf ("%d",&x);
        ++cnt[--x];
    }
    vector <int> c (k+1);
    rep (i,0,k-1) scanf ("%d",&c[i]);
    int j=0;
    vector <vector <int>> v (1);
    per (i,k-1,0) {
        if (c[i]>c[i+1]) j=0;
        while (cnt[i]--) {
           if (int (v[j].size ()==c[i])) j++;
           if (j==int (v.size ())) v.emplace_back ();
           v[j].pb (i);
        }
    }
    printf ("%d
",v.size ());
    for (auto it:v) {
        printf("%d",it.size ());
        for (auto i:it) printf (" %d",i+1);
        printf ("
");
    }
}

int main () {
    t=1;
    while (t--) {
        solve ();
    }
    return 0;
}

Problem E

首先要求每个都被一步攻击，那么需要每行或者每列都有一个车，然后刚好k对互相攻击，假设这个在每行放一个，那么这些车在列上面的分布一定是集中在n-k列之上，这样的话我们就得出答案，就是第二类斯特林数C（n,n-k）(n-k)!*2.

#include<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
#define rep(i,f_start,f_end) for (int i=f_start;i<=f_end;++i)
#define per(i,n,a) for (int i=n;i>=a;i--)
#define MT(x,i) memset(x,i,sizeof(x) )
#define rev(i,start,end) for (int i=start;i<end;i++)
#define inf 0x3f3f3f3f
#define mp(x,y) make_pair(x,y)
#define lowbit(x) (x&-x)
#define MOD 1000000007
#define exp 1e-8
#define N 1000005 
#define fi first 
#define se second
#define pb push_back
typedef long long ll;
const ll INF=0x3f3f3f3f3f3f3f3f;
typedef vector <int> VI;
typedef pair<int ,int> PII;
typedef pair<int ,PII> PIII;
ll gcd(ll a,ll b){return b==0?a:gcd(b,a%b);}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;};
void check_max (int &a,int b) { a=max (a,b);}
void check_min (int &a,int b) { a=min (a,b);}
inline int read() {
    char ch=getchar(); int x=0, f=1;
    while(ch<'0'||ch>'9') {
        if(ch=='-') f=-1;
        ch=getchar();
    } while('0'<=ch&&ch<='9') {
        x=x*10+ch-'0';
        ch=getchar();
    } return x*f;
}
const int mod=998244353;
const int maxn=2e5+10;
int t;
ll fac[maxn];
ll fast_pow (ll a,ll b) {
    ll ans=1;
    while (b) {
        if (b&1) ans=(ans*a)%mod;
        a=(a*a)%mod;
        b>>=1;
    }
    return ans;
}

ll C (ll n,ll m) {
    return fac[n]*fast_pow (fac[m]*fac[n-m]%mod,mod-2)%mod;
}

void solve () {
    ll ans=0,n,k;
    scanf ("%lld%lld",&n,&k);
    if (k>n-1) {
        printf ("0");
        return ;
    }
    fac[0]=fac[1]=1;
    ll m=n-k;
    rev (i,2,maxn) fac[i]=(fac[i-1]*i)%mod;
    rep (i,0,m) {
        if (i%2==0) ans+=C (m,i)*fast_pow (m-i,n)%mod;
        else ans=ans-C (m,i)*fast_pow (m-i,n)%mod+mod;
        ans%=mod;
    }
    ans= (ans*C (n,m))%mod;
    if (k) ans= (ans<<1)%mod;
    printf ("%lld
",ans);
}

int main () {
    // scanf ("%d",&t);
    t=1;
    while (t--) {
        solve ();
    }
    return 0;
}