A. Find Minimum Operations
暴力枚举
#include<iostream>
#include<algorithm>
#include<string>
#include<string.h>
#include<queue>
#include<deque>
#include<math.h>
#include<map>
#include<stack>
#include<set>
#include<vector>
#include<random>
using namespace std;
typedef long long ll;
void fio()
{ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
}
ll ksm(ll x,ll y)
{ll ans=1;while(y){if(y&1)ans*=x;x*=x;y>>=1;}return ans;
}
ll gcd(ll x,ll y)
{if(y==0)return x;elsereturn gcd(y,x%y);
}
int main()
{
fio();
ll t;
cin>>t;
while(t--)
{ll n,k;cin>>n>>k;if(k==1){cout<<n<<endl;continue;}else{ll ans=0;while(n){if(n<k){ans+=n;break;}else{ll cnt=1;for(ll i=1;i<=60;i++){cnt*=k;if(cnt>n){cnt/=k;break;}}n-=cnt;ans++;}}cout<<ans<<endl;}
}
}
B. Brightness Begins
通过分析可得答案使ans-sqrt（ans）>=n的最小ans值
这里要开srtl，否则怎么写都最多WA8
怎么说，一个sqrtl毁了我的梦（不是）
#include<iostream>
#include<algorithm>
#include<string>
#include<string.h>
#include<queue>
#include<deque>
#include<math.h>
#include<map>
#include<stack>
#include<set>
#include<vector>
#include<random>
using namespace std;
typedef unsigned long long ll;
void fio()
{ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
}
ll ksm(ll x,ll y)
{ll ans=1;while(y){if(y&1)ans*=x;x*=x;y>>=1;}return ans;
}
ll gcd(ll x,ll y)
{if(y==0)return x;elsereturn gcd(y,x%y);
}
int main()
{
fio();
ll t;
cin>>t;
while(t--)
{ll n,k;cin>>n;ll l=1,r=1844674407370955161;while(l<r){ll mid=(l+r)>>1;ll j=sqrtl(mid);if(mid-j>=n){r=mid;}elsel=mid+1;}cout<<r<<endl;
}
}
C. Bitwise Balancing
直接分析b，c，d对应位置地二进制关系，总共8种可能
随后直接走一遍二进制所有位即可，记得答案能小就尽量小
#include<iostream>
#include<algorithm>
#include<string>
#include<string.h>
#include<queue>
#include<deque>
#include<math.h>
#include<map>
#include<stack>
#include<set>
#include<vector>
#include<random>
using namespace std;
typedef long long ll;
void fio()
{ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
}
ll ksm(ll x, ll y)
{ll ans = 1;while (y){if (y & 1)ans *= x;x *= x;y >>= 1;}return ans;
}
ll gcd(ll x, ll y)
{if (y == 0)return x;elsereturn gcd(y, x % y);
}
int main()
{fio();ll t;cin >> t;while (t--){ll b, c, d;cin >> b >> c >> d;ll ans = 0;ll cnt = 1;ll g = 0;while (b||c){if (cnt > (ll)(1e18))break;ll e1 = (b & cnt);ll e2 = (c & cnt);if (e1)b -= cnt;if (e2)c -= cnt;ll op = (d & cnt);if (e1){if (e2){if (op){d -= cnt;}else{ans += cnt;}}else{if (op == 0){g = 1;}else{d -= cnt;}}}else{if (e2 == 0){if (op){ans += cnt;d -= cnt;}}else{if (op){g = 1;}}}cnt *= 2;}if (d > 0){ll cnt = 1;for (ll i = 0; i <= 62; i++){if (d & cnt){d -= cnt;ans += cnt;}if (d == 0)break;cnt *= 2;}}ll a = ksm(2, 61);//cout << a << endl;if (g==0&&ans<=a){cout << ans << endl;}elsecout << -1 << endl;}
}
