AtCoder Beginner Contest 242

A - T-shirt

Time Limit: 2 sec / Memory Limit: 1024 MB

Score : 100 points

Problem Statement
In a certain programming contest, T-shirts are awarded to participants according to the following rules.

All participants who ranked A-th or higher get a T-shirt.
Additionally, from the participants who ranked between (A+1)-th and B-th (inclusive), C participants chosen uniformly at random get a T-shirt.
There were 1000 participants in this contest, and all of them got different ranks.
Iroha-chan, who participated in this contest, ranked X-th.
Find the probability that she gets a T-shirt.

Constraints
All values in input are integers.
1≤A<B≤1000
1≤C≤B−A
1≤X≤1000
Input
Input is given from Standard Input in the following format:

A B C X
Output
Print the answer. Your output will be considered correct if the absolute or relative error from the judge’s answer is at most 10^−6
.

#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int>
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) {\
            For(j,m-1) cout<<a[i][j]<<' ';\
            cout<<a[i][m]<<endl;\
            }
#pragma comment(linker,"/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(a,b);
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{
  int x=0,f=1; char ch=getchar();
  while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
  while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
  return x*f;
} 
int main()
{
//  freopen("A.in","r",stdin);
//  freopen(".out","w",stdout);
  double a,b,c,x;
  cin>>a>>b>>c>>x;
  if(x<=a) cout<<"1"<<endl;
  else if(a+1<=x&&x<=b) {
    cout<<(double)(c)/(b-a)<<endl;
  }else cout<<0<<endl;
  
  return 0;
}

B - Minimize Ordering

You are given a string S. Find the lexicographically smallest string S’ obtained by permuting the characters of S.

#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int>
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) {\
            For(j,m-1) cout<<a[i][j]<<' ';\
            cout<<a[i][m]<<endl;\
            }
#pragma comment(linker,"/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(a,b);
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{
  int x=0,f=1; char ch=getchar();
  while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
  while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
  return x*f;
} 
int main()
{
//  freopen("B.in","r",stdin);
//  freopen(".out","w",stdout);
  char s[212345];
  cin>>(s+1);
  int n=strlen(s+1);
  sort(s+1,s+n+1);
  cout<<(s+1);
  
  
  return 0;
}

C - 1111gal password

Given an integer $N$, find the number of integers $X$ that satisfy all of the following conditions, modulo $998244353$.
$X$ is an $N$-digit positive integer.
Let $X_i$ be the digits of X from top to bottom. They satisfy all of the following:
$1\le X_i \le 9$ for all integers $1≤i≤N$
$∣X_i −X_{i+1}∣≤1$ for all integers $1≤i≤N−1$.

#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (998244353)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int>
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) {\
            For(j,m-1) cout<<a[i][j]<<' ';\
            cout<<a[i][m]<<endl;\
            }
#pragma comment(linker,"/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(a,b);
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{
  int x=0,f=1; char ch=getchar();
  while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
  while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
  return x*f;
} 
ll f[1123456][10];
int main()
{
//  freopen("C.in","r",stdin);
//  freopen(".out","w",stdout);
  int n=read();
  For(i,9)f[1][i]=1;
  For(i,n-1) {
    For(j,9) {
      upd(f[i+1][j],f[i][j]);
      if(j<9)upd(f[i+1][j+1],f[i][j]);
      if(j>1)upd(f[i+1][j-1],f[i][j]);
    }
  }
  ll ans=0;
  For(j,9) upd(ans,f[n][j]);cout<<ans<<endl;  
  
  return 0;
}

D - ABC Transform

You are given a string $S$ consisting of $A, B, C$.

Let $S(0):=S$. For $i=1,2,3,…,$ let $S(i)$ be the result of simultaneously replacing the characters of $S(i−1)$ as follows: $A → BC, B → CA, C → AB$.

Answer $Q$ queries. The $i$-th query is as follows.

Print the $k$-th character from the beginning of $S(t_i)$

#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (998244353)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int>
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) {\
            For(j,m-1) cout<<a[i][j]<<' ';\
            cout<<a[i][m]<<endl;\
            }
#pragma comment(linker,"/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(a,b);
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{
  int x=0,f=1; char ch=getchar();
  while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
  while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
  return x*f;
} 
char s[112345];
int main()
{
//  freopen("D.in","r",stdin);
//  freopen(".out","w",stdout);
  cin>>(s+1);
  int n=strlen(s+1);
  int q=read();
  For(i,q) {
    ll t,k;cin>>t>>k;
    ll p=0;
    while(t>0 && k>1) {
      p=p+1+(k%2==0);
      p%=3ll;
      k=(k+1)/2;
      --t;
    }
    if(t>0) p=(p+t%3)%3;
    char cc=((s[k]-'A')+p)%3+'A';
    cout<<cc<<endl;
  }
  
  
  return 0;
}

E - (∀x∀)

Solve the following problem for T test cases.

Given an integer N and a string S, find the number of strings X that satisfy all of the conditions below, modulo 998244353.

X is a string of length N consisting of uppercase English letters.
X is a palindrome.
X≤S in lexicographical order.
That is, X=S or X is lexicographically smaller than S.

#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (998244353)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int>
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) {\
            For(j,m-1) cout<<a[i][j]<<' ';\
            cout<<a[i][m]<<endl;\
            }
#pragma comment(linker,"/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(a,b);
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{
  int x=0,f=1; char ch=getchar();
  while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
  while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
  return x*f;
} 
char s[1123456];
char s2[1123456];
ll p26[1123456];
int main()
{
//  freopen("E.in","r",stdin);
//  freopen(".out","w",stdout);
  p26[0]=1;
  For(i,1e6) p26[i]=p26[i-1]*26ll%F;
  
  int T=read();
  
  while(T--) {
    ll ans=0;
    int n=read();
    cin>>(s+1);    
    int m=(n+1)/2;
    For(i,m) {
      int p=s[i]-'A';
      upd(ans,mul(p,p26[m-i]));
    }
    For(i,m) s2[i]=s[i];
    Fork(i,m+1,n) s2[i]=s2[n-i+1];
    s2[0]=s2[n+1]=0;
//    cout<<(s2+1)<<endl;
//    cout<<(s+1)<<endl;
    if (strcmp(s2+1,s+1)<=0) upd(ans,1);
    cout<<ans<<endl;
  }
  
  
  return 0;
}

F - Black and White Rooks

Time Limit: 2 sec / Memory Limit: 1024 MB

Score : 500 points

Problem Statement
在一个$N*M$的矩阵中放$B$个黑石子和$W$个白石子。

要求：
所有石子放在一个格子里
一个格子最多放一个石子。
同行或同列，没有同色石子。

求方案数 modulo 998244353.

同色石子无区别。

Constraints
1≤N,M≤50
1≤B,W≤2500
B+W≤N×M
All values in input are integers.
Input
Input is given from Standard Input in the following format:

N M B W
Output
Print the count modulo 998244353.

Sample Input 1
Copy
2 2 1 1
Sample Output 1
Copy
4

容斥原理：
答案$=\sum_{i=1}^n\sum_{j=1}^m\sum_{k=1}^{n-i}\sum_{l=1}^{m-j}(-1)^{n+m-i-j-k-l}*C(n,i)C(n-i,k)C(m,j)C(m-j,l)C(ij,w)C(kl,b)$

#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (998244353)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int>
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) {\
                        For(j,m-1) cout<<a[i][j]<<' ';\
                        cout<<a[i][m]<<endl;\
                        }
#pragma comment(linker,"/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(a,b);
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{
    int x=0,f=1; char ch=getchar();
    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
    return x*f;
} 
#define MAXN (101015)
ll inj[MAXN],jie[MAXN];
inline ll C(int a,int b) {
    return (ll)jie[a]*inj[b]%F*inj[a-b]%F;
}
ll p=F;
inline int pow2(int a,ll b)  //a^b mod p 
{  
    if (b==0) return 1%p;  
    int c=pow2(a,b/2);  
    c=(ll)c*c%p;  
    if (b&1) c=(ll)c*a%p;  
    return c;  
}  
void pre(int n) {
    jie[0]=1;For(i,n) jie[i]=mul(jie[i-1],i);
    inj[0]=inj[1]=1;Fork(i,2,n) inj[i]=(F-(F/i))*inj[F%i]%F;
    For(i,n) inj[i]=mul(inj[i],inj[i-1]);  
} 
ll n,m,b,w;
int main()
{
//  freopen("E.in","r",stdin);
//  freopen(".out","w",stdout);
    pre(101010);
    ll ans=0;
    cin>>n>>m>>b>>w;
    For(i,n) For(j,m) For(k,n-i) For(l,m-j) if(k*l>=b &&i*j>=w){
        ll p=mul(C(n,i),C(m,j));
        p=mul(p,C(n-i,k));
        p=mul(p,C(m-j,l));
        p=mul(p,C(k*l,b));
        p=mul(p,C(i*j,w));
        if((n+m-i-j-k-l)%2) p=sub(0,p);
        upd(ans,p);
    }
    cout<<ans<<endl;
    return 0;
}

G - Range Pairing Query

N people numbered $1,2,…,N$ are standing in a row. Person i wears Color $A_i$
Answer $Q$

You are given integers $l$ and $r$. Considering only Person $l,l+1,…,r,$

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cstring>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define MAXN (1123456+10)
#define MAXM MAXN
int n,m,a[MAXN];
int b[MAXN]={0};
long long Ti2=0;
struct commend
{
  int l,r,no;
  commend(){}
  commend(int _l,int _r,int _no):l(_l),r(_r),no(_no){}
  friend bool operator<(commend a,commend b){return int(a.l/sqrt(n))<int(b.l/sqrt(n))||(int(a.l/sqrt(n))==int(b.l/sqrt(n)))&&a.r<b.r;  }
}p1[MAXM],p2[MAXM];
long long ans[MAXN];
long long gcd(long long a,long long b){if (!b) return a;return gcd(b,a%b);}
int main()
{
//  freopen("bzoj2038.in","r",stdin);
//  freopen(".out","w",stdout);
  scanf("%d",&n);
  For(i,n) scanf("%d",&a[i]);
  scanf("%d",&m);
  For(i,m) scanf("%d%d",&p1[i].l,&p1[i].r),p1[i].no=i;
  memcpy(p2,p1,sizeof(p1));
  sort(p2+1,p2+1+m);p2[0].l=1,p2[0].r=0;
  For(i,m)
  {
    if (p2[i-1].l<p2[i].l) Fork(j,p2[i-1].l,p2[i].l-1) Ti2-=(long long)b[a[j]]/2,b[a[j]]--,Ti2+=(long long)b[a[j]]/2;
    if (p2[i].l<p2[i-1].l) Fork(j,p2[i].l,p2[i-1].l-1) Ti2-=(long long)b[a[j]]/2,b[a[j]]++,Ti2+=(long long)b[a[j]]/2;
    if (p2[i-1].r<p2[i].r) Fork(j,p2[i-1].r+1,p2[i].r) Ti2-=(long long)b[a[j]]/2,b[a[j]]++,Ti2+=(long long)b[a[j]]/2;
    if (p2[i].r<p2[i-1].r) Fork(j,p2[i].r+1,p2[i-1].r) Ti2-=(long long)b[a[j]]/2,b[a[j]]--,Ti2+=(long long)b[a[j]]/2;
    ans[p2[i].no]=Ti2;
  }
  For(i,m)
    cout<<ans[i]<<endl;
  return 0;
}