Codeforces Round 1011 (Div. 2)（A~D,补了E）
比赛链接：https://codeforces.com/contest/2085
A.Serval and String Theory
思路：我看了快4分钟才明白题意，显然字符串的字符全相等或者k==0时，字符串不符合题意即无解，否则就是对的
#include<iostream>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<algorithm>
#include<deque>
#include<cctype>
#include<string.h>
#include<math.h>
#include<time.h>
#include<random>
#include<functional>
#include<stack>
#include<unordered_map>
#include<string>
#define ll long long
#define lowbit(x) (x & -x)
#define endl "\n"// 交互题记得删除
// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>
// using namespace __gnu_pbds;
using namespace std;
mt19937 rnd(time(0));
const ll mod = 998244353;
//const ll p=rnd()%mod;
#define F first
#define S second
// tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update>;find_by_order(k)，找第k位（从小到大）的数字，order_of_key(x)，x的排名
ll ksm(ll x, ll y)
{
ll ans = 1;
x %= mod;
while (y)
{
if (y & 1)
{
ans = ans * x % mod;
}
x = x * x % mod;
y >>= 1;
}
return ans;
}
ll gcd(ll x, ll y)
{
if (y == 0)
return x;
else
return gcd(y, x % y);
}
inline ll read()
{
register ll x = 0, f = 1;
char c = getchar();
while (c < '0' || c>'9')
{
if (c == '-') f = -1;
c = getchar();
}
while (c >= '0' && c <= '9')
{
x = (x << 3) + (x << 1) + (c ^ 48); //等价于x*10+c-48,使用位运算加速
c = getchar();
}
return x * f;
}
void fio()
{
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
struct s
{
ll l,r;
friend bool operator<(const s &a, const s& b)
{
return a.l<b.l;
};
};
int main()
{
fio();
ll t;
cin >> t;
while (t--)
{
ll n,k;
cin>>n>>k;
string f;
cin>>f;
ll cnt=0;
for(ll i=0;i<n;i++)
{
cnt+=(f[i]==f[0]);
}
ll pd=0;
for(ll i=0;i<n;i++)
{
if(f[i]<f[n-i-1])
{
pd=1;
break;
}
else if(f[i]>f[n-i-1])
{
pd=2;
break;
}
}
if(cnt==n||(k==0&&(pd==2||pd==0)))
{
cout<<"NO"<<endl;
}
else cout<<"YES"<<endl;
}
return 0;
}
B.Serval and Final MEX
思路：全0就分成一半一半，然后统一合并。全为大于0的数字就直接全合成就好了。有0和有大于0的数字，我是这样想的。
如果最后两位有至少一个0出现，那么他们两个合并下，然后去遍历前面的数(n-2个数)看下前面的数否全为大于0的数，如果前面全大于0则不合成，最后大合成就好
否则，最后两个数不合成，前面n-2 个数合成下，最后再大合成即可
#include<iostream>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<algorithm>
#include<deque>
#include<cctype>
#include<string.h>
#include<math.h>
#include<time.h>
#include<random>
#include<functional>
#include<stack>
#include<unordered_map>
#include<string>
#define ll long long
#define lowbit(x) (x & -x)
#define endl "\n"// 交互题记得删除
// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>
// using namespace __gnu_pbds;
using namespace std;
mt19937 rnd(time(0));
const ll mod = 998244353;
//const ll p=rnd()%mod;
#define F first
#define S second
// tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update>;find_by_order(k)，找第k位（从小到大）的数字，order_of_key(x)，x的排名
ll ksm(ll x, ll y)
{
ll ans = 1;
x %= mod;
while (y)
{
if (y & 1)
{
ans = ans * x % mod;
}
x = x * x % mod;
y >>= 1;
}
return ans;
}
ll gcd(ll x, ll y)
{
if (y == 0)
return x;
else
return gcd(y, x % y);
}
inline ll read()
{
register ll x = 0, f = 1;
char c = getchar();
while (c < '0' || c>'9')
{
if (c == '-') f = -1;
c = getchar();
}
while (c >= '0' && c <= '9')
{
x = (x << 3) + (x << 1) + (c ^ 48); //等价于x*10+c-48,使用位运算加速
c = getchar();
}
return x * f;
}
void fio()
{
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
struct s
{
ll l,r;
friend bool operator<(const s &a, const s& b)
{
return a.l<b.l;
};
};
int main()
{
fio();
ll t;
cin >> t;
while (t--)
{
ll n;
cin>>n;
vector<ll>a(n+5);
ll cnt=0;
ll wz;
for(ll i=1;i<=n;i++)
{
cin>>a[i];
cnt+=(a[i]==0);
}
if(cnt==0)
{
cout<<1<<endl;
cout<<1<<" "<<n<<endl;
}
else if(cnt!=n)
{
if(a[n-1]==0||a[n]==0)
{
ll pd=0;
ll f=0;
for(ll i=1;i<=n-2;i++)
{
if(a[i]>0)f++;
}
if(f!=n-2)
{
cout<<3<<endl;
cout<<n-1<<" "<<n<<endl;
cout<<1<<" "<<n-2<<endl;
cout<<1<<" "<<2<<endl;
}
else
{
cout<<2<<endl;
cout<<n-1<<" "<<n<<endl;
cout<<1<<" "<<n-1<<endl;
}
}
else
{
cout<<2<<endl;
cout<<1<<" "<<n-2<<endl;
cout<<1<<" "<<3<<endl;
}
}
else
{
cout<<3<<endl;
cout<<n-1<<" "<<n<<endl;
cout<<1<<" "<<n-2<<endl;
cout<<1<<" "<<2<<endl;
}
}
return 0;
}
C.Serval and The Formula
思路：其实仔细写写发现就是去尽量消除x和y在某个二进制位同为1的情况，发现能消除的方法就是找更低的二进制位为（0和1）的情况，然后不断进位即可。写的时间复杂度就60*60吧，其实还有O(1)办法，应该是这个吧，(1ll<<30)-max(x,y).
#include<iostream>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<algorithm>
#include<deque>
#include<cctype>
#include<string.h>
#include<math.h>
#include<time.h>
#include<random>
#include<functional>
#include<stack>
#include<unordered_map>
#include<string>
#define ll long long
#define lowbit(x) (x & -x)
#define endl "\n"// 交互题记得删除
// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>
// using namespace __gnu_pbds;
using namespace std;
mt19937 rnd(time(0));
const ll mod = 998244353;
//const ll p=rnd()%mod;
#define F first
#define S second
// tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update>;find_by_order(k)，找第k位（从小到大）的数字，order_of_key(x)，x的排名
ll ksm(ll x, ll y)
{
ll ans = 1;
x %= mod;
while (y)
{
if (y & 1)
{
ans = ans * x % mod;
}
x = x * x % mod;
y >>= 1;
}
return ans;
}
ll gcd(ll x, ll y)
{
if (y == 0)
return x;
else
return gcd(y, x % y);
}
inline ll read()
{
register ll x = 0, f = 1;
char c = getchar();
while (c < '0' || c>'9')
{
if (c == '-') f = -1;
c = getchar();
}
while (c >= '0' && c <= '9')
{
x = (x << 3) + (x << 1) + (c ^ 48); //等价于x*10+c-48,使用位运算加速
c = getchar();
}
return x * f;
}
void fio()
{
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
struct s
{
ll l,r;
friend bool operator<(const s &a, const s& b)
{
return a.l<b.l;
};
};
int main()
{
fio();
ll t;
cin >> t;
while (t--)
{
ll x,y;
cin>>x>>y;
//cout<<
ll k=0;
ll ok=0;
for(ll i=0;i<=60;i++)
{
ll z=(1ll<<i);
if((x&z)&&(y&z))
{
while((x&z)&&(y&z))
{
ll pd=0;
for(ll j=i-1;j>=0;j--)
{
ll o=1ll<<j;
if(((x&o)==0&&(y&o)>0)||((y&o)==0&&(x&o)>0))
{
pd=1;
x+=o;
y+=o;
k+=o;
break;
}
}
if(pd==0)break;
}
if(((x&z)&&(y&z))&&i>=55)
{
ok=1;
break;
}
else if((x&z)&&(y&z))
{
x+=z;
y+=z;
k+=z;
}
}
}
if(ok||k>1e18)cout<<-1<<endl;
else cout<<k<<endl;
}
return 0;
}
//100100100100100100100
// 10010010010010010100
D.Serval and Kaitenzushi Buffet
思路：反悔贪心，每次看吃的时间够不够，够得话就拿走吃的，否则就看要不要最小的换下即可
#include<iostream>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<algorithm>
#include<deque>
#include<cctype>
#include<string.h>
#include<math.h>
#include<time.h>
#include<random>
#include<functional>
#include<stack>
#include<unordered_map>
#include<string>
#define ll long long
#define lowbit(x) (x & -x)
#define endl "\n"// 交互题记得删除
// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>
// using namespace __gnu_pbds;
using namespace std;
mt19937 rnd(time(0));
const ll mod = 998244353;
//const ll p=rnd()%mod;
#define F first
#define S second
// tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update>;find_by_order(k)，找第k位（从小到大）的数字，order_of_key(x)，x的排名
ll ksm(ll x, ll y)
{
ll ans = 1;
x %= mod;
while (y)
{
if (y & 1)
{
ans = ans * x % mod;
}
x = x * x % mod;
y >>= 1;
}
return ans;
}
ll gcd(ll x, ll y)
{
if (y == 0)
return x;
else
return gcd(y, x % y);
}
inline ll read()
{
register ll x = 0, f = 1;
char c = getchar();
while (c < '0' || c>'9')
{
if (c == '-') f = -1;
c = getchar();
}
while (c >= '0' && c <= '9')
{
x = (x << 3) + (x << 1) + (c ^ 48); //等价于x*10+c-48,使用位运算加速
c = getchar();
}
return x * f;
}
void fio()
{
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
struct s
{
ll v, id;
friend bool operator<(const s& a, const s& b)
{
return a.v > b.v;
};
};
int main()
{
fio();
ll t;
cin >> t;
while (t--)
{
ll n, k;
cin >> n >> k;
vector<ll>a(n + 5);
for (ll i = 1; i <= n; i++)
{
cin >> a[i];
}
ll cnt = k;
priority_queue<s>f;
ll ans = 0;
ll sum = 0;
for (ll i = n - k; i >= 1; i--)
{
if (cnt >= k)
{
f.push({ a[i],i });
sum += a[i];
cnt -= k;
}
else
{
ll v = f.top().v;
if (v >= a[i])
{
cnt++;
continue;
}
else
{
sum -= v;
f.pop();
f.push({ a[i],i });
cnt++;
sum += a[i];
}
}
ans = max(ans, sum);
}
cout << ans << endl;
}
return 0;
}
//100100100100100100100
// 10010010010010010100
E.Serval and Modulo
思路：赛后补的。看了jiangly的代码。对于正确的{a[i],b[i]}对，其实(a[i]-b[i])一定是k的倍数，除了a数组等于b数组情况。
所以我们可以把所有(a[i]-b[i])加起来即可，然后分解因数去检测下就可以做出来，实际因数不多，其实不等于nsqrt（n）logn，时间复杂度远比这个小,因为因数其实很少，不是因数的时间复杂度上算加法
#include<iostream>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<algorithm>
#include<deque>
#include<cctype>
#include<string.h>
#include<math.h>
#include<time.h>
#include<random>
#include<functional>
#include<stack>
#include<unordered_map>
#include<string>
#define ll long long
#define lowbit(x) (x & -x)
#define endl "\n"// 交互题记得删除
// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>
// using namespace __gnu_pbds;
using namespace std;
mt19937 rnd(time(0));
const ll mod = 998244353;
//const ll p=rnd()%mod;
#define F first
#define S second
// tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update>;find_by_order(k)，找第k位（从小到大）的数字，order_of_key(x)，x的排名
ll ksm(ll x, ll y)
{
ll ans = 1;
x %= mod;
while (y)
{
if (y & 1)
{
ans = ans * x % mod;
}
x = x * x % mod;
y >>= 1;
}
return ans;
}
ll gcd(ll x, ll y)
{
if (y == 0)
return x;
else
return gcd(y, x % y);
}
inline ll read()
{
register ll x = 0, f = 1;
char c = getchar();
while (c < '0' || c>'9')
{
if (c == '-') f = -1;
c = getchar();
}
while (c >= '0' && c <= '9')
{
x = (x << 3) + (x << 1) + (c ^ 48); //等价于x*10+c-48,使用位运算加速
c = getchar();
}
return x * f;
}
void fio()
{
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
struct s
{
ll v, id;
friend bool operator<(const s& a, const s& b)
{
return a.v > b.v;
};
};
int main()
{
fio();
ll t;
cin >> t;
while (t--)
{
ll n;
cin>>n;
//如果有答案，那么差值一定包含这个答案的倍数
vector<ll>a(n+5,0),b(n+5,0);
for(ll i=1;i<=n;i++)cin>>a[i];
for(ll i=1;i<=n;i++)cin>>b[i];
sort(a.begin()+1,a.begin()+1+n);
sort(b.begin()+1,b.begin()+1+n);
ll f=0;
for(ll i=1;i<=n;i++)
{
f+=a[i]-b[i];
}
if(a==b)
{
cout<<(ll)1e9<<endl;
continue;
}
if(f<=0)
{
cout<<-1<<endl;
continue;
}
function<ll(ll)>ck=[&](ll x)
{
if(x>(ll)1e9)return 0;
vector<ll>f(n+5,0);
f=a;
for(ll i=1;i<=n;i++)
{
f[i]%=x;
}
sort(f.begin()+1,f.begin()+1+n);
if(f==b)
return 1;
else return 0;
};
ll ans=-1;
for(ll i=1;i*i<=f;i++)
{
if(f%i==0)
{
if(ck(i))
{
ans=i;
break;
}
if(ck(f/i))
{
ans=f/i;
break;
}
}
}
cout<<ans<<endl;
}
}
//100100100100100100100
// 10010010010010010100
浙公网安备 33010602011771号