AtCoder Regular Contest 154 题解

Posted 2023-01-24 nike0good

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个数相同位置的数字，要求使它们乘积最小

让其中一个数最小，另一个数最大。

#include<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;
 
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<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);
	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$以上是关于AtCoder\; Regular\; Contest\; 154\; 题解的主要内容，如果未能解决你的问题，请参考以下文章$

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