AtCoder Regular Contest 154 题解

A - Swap Digit

给$2$个长度均为$n$的十进制数，你可以任意次交换$2$个相同位置的数字，要求使它们乘积最小

让其中一个数最小，另一个数最大。

#include 
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;
} 
string a,b;
int main()
{
//	freopen("A.in","r",stdin);
//	freopen(".out","w",stdout);
	int n=read();
	cin>>a>>b;
	ll p=0,q=0;
	Rep(i,n) {
		if(a[i]<b[i]) p=(p*10+a[i]-'0')%F,q=(q*10+b[i]-'0')%F;
		else {
			p=(p*10+b[i]-'0')%F,q=(q*10+a[i]-'0')%F;
		}
		
	}
	cout<<mul(p,q);
	return 0;
}

B - New Place

给$2$个长度为$n$的串，每次可以把第一个串的第一个字符塞进这个字符串任意位置，问把这两个串变相同的最小次数。无解$-1$。

有解当且仅当各个字符在2个字符串中出现次数相同
贪心匹配第一个字符串中的后缀

#include 
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);
	int n=read();
	string s,t;
	cin>>s>>t;
	int c[256]={};
	Rep(i,n) c[s[i]]++,c[t[i]]--;
	Fork(i,'a','z') if(c[i]){
		puts("-1");return 0;
	}
	int p=n-1,ans=0;
	RepD(i,n-1) {
		while(p>=0 && s[i]!=t[p]) --p;
		if(p<0) break;
		--p,++ans;
	}cout<<n-ans;
	
	return 0;
}

C - Roller

You are given sequences of positive integers of length $A,B$
You can repeat the following operation any number of times (possibly zero).
Choose an integer $i$ such that $1\le i\le N$ and A_i :=A_i+1​
.
Here, regard $A_{N+1}$ as $A_1$

Determine whether it is possible to make $A$ equal $B$.

$A$有一个数不在了就可以循环平移。
最后要么$A$和$B$一开始就一样，要么$A$平移后能与$B$的段一一对应。

#include 
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;
} 
#define MAXN (5000+10)
int a[MAXN],b[MAXN],c[MAXN];
bool work(){
	int n=read();
	Rep(i,n) a[i]=read();
	Rep(i,n) b[i]=read();
	//case 1
	bool fl=0;
	Rep(i,n) if(a[i]!=b[i]) fl=1;
	if(!fl) return 1;
	//case 2
	int p=0,q=0;
	Rep(i,n) if(a[i]!=(a[(i+1)%n])) ++p;
	Rep(i,n) if(b[i]!=(b[(i+1)%n])) ++q;
	if(p==n&&q==n) return 0;
	if(p<q) return 0;
	//case 3
	
	Rep(i,n) c[i]=b[i];
	int m=unique(c,c+n)-c;
//	Rep(i,m) cout<
	if(m>=2 &&c[m-1]==c[0]) --m;
	
	Rep(t,n) {
		int p=0;
		Rep(len,n) {
			int i=(t+len)%n;
			if(a[i]==c[p]) {
				++p;
			}
			else continue;			
		}	
		if(p>=m) return 1;
	}
	
	return 0;
	
}
int main()
{
//	freopen("C.in","r",stdin);
//	freopen(".out","w",stdout);
	int T=read();
	while(T--) puts(work()?"Yes":"No");
	
	
	return 0;
}

D - A + B > C ?

PCT has a permutation $P_1,\dots ,P_N)$. You are only informed of $N\le 2000$.

You can ask him at most $25000$ questions of the following form.

Specify a triple of integers $(i,j,k)$ such that $1\le i,j,k\le N$ and ask whether $P_i+P_j<P_k$
($i,j,k$ can be equal)

$P_x+P_x<=P_y(y\ne x)$,则KaTeX parse error: Double subscript at position 7: P_y<_P_̲x，所以可以通过遍历$y$来找出比$x$小的数，如果找出则$y$更新$x$，就能找出$i{d}_1(P_{i{d}_1}=1)$
于是排列中的不同数差绝对值至少为1，$a<b$等价于$a+1<=b$，剩下的数可以用归并排序，比较次数$O(nlogn)$
注意为了避免最坏情况要用stable_sort代替sort
https://legacy.cplusplus.com/reference/algorithm/stable_sort/?kw=stable_sort

#include 
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;
} 
#define MAXN (2010)
int n,id[MAXN]={},cnt=0;
bool get() {
	string s;
	cin>>s;
	return s[0]=='Y';
}
bool ask(int a,int b,int c) {
	cout<<"? "<<a<<' '<<b<<' '<<c<<endl;
	fflush(stdout);
	return get();
}
int cmp(int i,int j) {
	return ask(i,id[1],j)^1;
}
int main()
{
//	freopen("D.in","r",stdin);
//	freopen(".out","w",stdout);
	cin>>n;
	For(i,n) id[i]=i;
	Fork(i,2,n) if(!ask(i,i,id[1])){
		swap(id[1],id[i]);
	}
	if(n>1)
		stable_sort(id+2,id+n+1,cmp);
	int p[MAXN];
	For(i,n) p[id[i]]=i;
	cout<<"!";
	For(i,n) cout<<' '<<p[i];cout<<endl;
	fflush(stdout);
	return 0;
}