8.30

• 二分图匹配
• 树的直径
• 树的重心
• gcd 和 exgcd
• 乘法逆元

照常：

• noip模拟：火柴棒等式/机器翻译/P1079 Vigenère 密码
• noip DP

---

火柴棒等式

#define N 1111 

int biao[]={6,2,5,5,4,5,6,3,7,6};
int n,ans;

int get_num(int x){
	int ans=0;
	if(x==0)ans=biao[0];//特判x==0的情况！
	while(x){
		ans+=biao[x%10];
		x/=10;
	}
	return ans;
}

int main(){
	freopen("huochai.txt","r",stdin);
	rd(n);
	//printf("%d
",get_num(0));
	rep(i,0,N){
		rep(j,0,N){
			if(get_num(i)+get_num(j)+get_num(i+j)+4==n){
				ans++;
				//printf("%d %d %d %d %d %d 
",i,get_num(i),j,get_num(j),i+j,get_num(i+j));
			}
		}
	}
	printf("%d",ans);
	return 0;
}

---

机器翻译

先进先出，队列裸题，
注意queue的size好像只能返回01……还是开一个变量来记录数组的大小吧。

#define N 1010

int m,n,ans,cnt;
int a[N];
queue<int>q;
map<int,int>mp;

int main(){
	rd(m),rd(n);
	rep(i,1,n){
		rd(a[i]);
		if(cnt<m){
			if(mp[a[i]])continue;
			++cnt;
			if(mp[a[i]]==0)
				ans++; 
			q.push(a[i]);
			mp[a[i]]=1;
		}
		else if(cnt>=m){
			if(mp[a[i]])continue;///////////题目条件，你怎么能不写呢？ 
			int u=q.front();
			mp[u]=0;
			q.pop();
			if(mp[a[i]]==0)ans++;
			q.push(a[i]);
			mp[a[i]]=1;
		} 
	}
	printf("%d",ans);
	return 0;
}

P1079 Vigenère 密码

真是累死宝宝了…… 码了接近一小时，得了10pts哈哈哈哈哈
不过学到了新技巧：如何快速将大写转化为小写？
x&31就好啦

/*
map<char,int>mp;
string k,mi;

inline void init(){
	mp['a']=mp['A']=0;
	mp['b']=mp['B']=1;
	mp['c']=mp['C']=2;
	mp['d']=mp['D']=3;
	mp['e']=mp['E']=4;
	mp['f']=mp['F']=5;
	mp['g']=mp['G']=6;
	mp['h']=mp['H']=7;
	mp['i']=mp['I']=8;
	mp['j']=mp['J']=9;
	mp['k']=mp['K']=10;
	mp['l']=mp['L']=11;
	mp['m']=mp['M']=12;
	mp['n']=mp['N']=13;
	mp['o']=mp['O']=14;
	mp['p']=mp['P']=15;
	mp['q']=mp['Q']=16;
	mp['r']=mp['R']=17;
	mp['s']=mp['S']=18;
	mp['t']=mp['T']=19;
	mp['u']=mp['U']=20;
	mp['v']=mp['V']=21;
	mp['w']=mp['W']=22;
	mp['x']=mp['X']=23;
	mp['y']=mp['Y']=24;
	mp['z']=mp['Z']=25;
}

int main(){
	init();
	//printf("%c",'f'-mp['b']);
	cin>>k>>mi;
	int lenk=k.length(),lenmi=mi.length();
	if(lenk<lenmi){
		int times=lenmi/lenk;
		string yuan=k;
		rep(i,2,times)
			k+=yuan;
		rep(i,times*lenk+1,lenmi)
			k+=k[i%(lenk+1)];
	}
	cout<<k<<endl;
	for(int i=0;i<lenmi;++i){
		if(mp[mi[i]]-mp[k[i]]<0)
			printf("%c",mi[i]+(26-mp[k[i]]));
		else 
			printf("%c",mi[i]-mp[k[i]]);
	}
	return 0;
}*/


int main(){
	//freopen("A.txt","r",stdin);
	string k,c;
	cin>>k>>c;
	for(int i=0;i<c.length();++i){
		int now=(k[i%k.length()]&31)-1;
		if(((c[i]&31)-now)>0)
			c[i]-=now;
		else
			c[i]-=now-26;
	}
	cout<<c<<endl;
	return 0;
}