题目链接
A. The Man who became a God
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| #include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1010;
int t,n,k;
int a[N];
int mins[N];
int main()
{
cin >> t;
while(t--)
{
cin >> n >> k;
for(int i = 1;i <= n;++i)
cin >> a[i];
for(int i = 1;i < n;++i)
mins[i] = abs(a[i + 1] - a[i]);
sort(mins + 1,mins + n);
ll ans = 0;
k--;
for(int i = n - 1;i >= 1;--i)
{
if(k)
k--;
else
ans += mins[i];
}
cout << ans << '\n';
}
}
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B. Hamon Odyssey
显然取并后最小结果不为0的情况有且只有一种，结果为0则贪心计算最大方案数
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| #include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 2e5 + 10;
int n,t;
ll a[N];
ll mins = 0;
int main()
{
cin >> t;
while(t--)
{
cin >> n;
for(int i = 1;i <= n;++i)
{
cin >> a[i];
}
mins = a[1];
for(int i = 2;i <= n;++i)
mins &= a[i];
if(mins != 0)
{
cout << 1 << '\n';
}else
{
int cnt = 0;
ll temp = a[1];
for(int i = 2;i <= n;++i)
{
if(temp == 0)
{
cnt++;
temp = a[i];
}else
temp &= a[i];
}
if(temp == 0)
cnt++;
cout << cnt << endl;
}
}
}
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C. Vampiric Powers, anyone?
实际上是求一段最大子段异或和，由于很小可以直接记录所有可能值取最大
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| #include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5 + 10;
int a[N];
int n,t;
bool vis[300];
int main()
{
ios::sync_with_stdio(0);
cin.tie(0),cout.tie(0);
cin >> t;
while(t--)
{
cin >> n;
ll ans = 0;
memset(vis,0,sizeof(vis));
for(int i = 1;i <= n;++i)
{
cin >> a[i];
}
vis[0] = 1;
ll temp = 0;
for(int i = 1;i <= n;++i)
{
temp ^= a[i];
for(int j = 0;j < 256;++j)
{
if(vis[j])
ans = max(ans,temp ^ j);
}
vis[temp] = 1;
}
cout << ans << endl;
}
}
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