\(A - Legendary Players\)
map<char ,int >mp;
void solve(){
string s;
cin>>s;
cout<<mp[s[0]]<<'\n';
//puts(ans>0?"YES":"NO");
//puts(ans>0?"Yes":"No");
}
signed main(){
mp['t']=3858;
//ksun48
mp['k']=3679;
// Benq
mp['B']=3658;
// Um_nik
mp['U']=3648;
// apiad
mp['a']=3638;
// Stonefeang
mp['S']=3630;
// ecnerwala
mp['e']=3613;
// mnbvmar
mp['m']=3555;
// newbiedmy
mp['n']=3516;
// semiexp
mp['s']=3481;
int t=1;
while(t--){
solve();
}
}
\(B - Measure\)
理解了半天题意,然后就按照题目的意思暴力模拟,我现在也没懂他在干什么。
void solve(){
int n=read();
for(int i=0;i<=n;i++){
if(i==0){
cout<<1;
continue;
}
int ans=10;
for(int j=1;j<=9;j++){
if(n%j==0&&i%(n/j)==0){
ans=min(ans,j);
}
}
if(ans==10)cout<<'-';
else cout<<ans;
}
//puts(ans>0?"YES":"NO");
//puts(ans>0?"Yes":"No");
}
\(C - False Hope\)
\(D - Minimum Width\)
这题也是显然的二分答案,每次 \(check\) 的时候 \(O(n)\) 模拟。
int a[N],n,k;
bool check(int x) {
int now=a[1],h=1;
for(int i=2;i<=n;i++){
if(now+a[i]+1<=x){
now+=a[i]+1;
}else{
now=a[i];
h++;
}
}
return h<=k;
}
int bs1(int l, int r){ //左偏二分
while (l < r){
int mid = (l + r )>> 1;
if (check(mid)) r = mid;
else l = mid + 1;
}
return l;
}
void solve(){
n=read(),k=read();
int sum=0,maxx=0;
for(int i=1;i<=n;i++){
a[i]=read();
maxx=max(a[i],maxx);
sum+=a[i];
}
cout<<bs1(maxx,sum*2)<<'\n';
//puts(ans>0?"YES":"NO");
//puts(ans>0?"Yes":"No");
}
\(E - Bus Stops\)
显然对于所有间隔的 \(lcm\) ,余数相同的时候答案应该相同。看到间隔的范围很小,可以自然的想到预处理出 \(lcm\) 内的所有答案,然后就可以 \(O(1)\) 的回答询问。
int n,m,a[N],b[N];
int ans[N];
int gcd(int a, int b){
return b ? gcd(b, a % b) : a;
}
void yuchuli(){
m=1;
for(int i=1;i<=8;i++){
m=m*i/gcd(m,i);
}
for(int i=0;i<=m;i++){
int x=i;
for(int j=1;j<n;j++){
if(x%a[j])x+=a[j]-x%a[j];
x+=b[j];
}
ans[i]=x;
}
}
void solve(){
n=read();
int s=read(),t=read();
for(int i=1;i<n;i++){
a[i]=read();
b[i]=read();
}
yuchuli();
int q=read();
while(q--){
int x=read();
x+=s;
x+=ans[x%m]-x%m;
x+=t;
cout<<x<<'\n';
}
//puts(ans>0?"YES":"NO");
//puts(ans>0?"Yes":"No");
}
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