Codeforces Round #661 (Div. 3)
A : R e m o v e A:Remove A:Remove S m a l l e s t Smallest Smallest
思路：
排序后，若有两数差大于一的，则输出 “ N O ” “NO” “NO”，反则输出 “ Y E S ” “YES” “YES”
参考代码：
#include p;
//priority_queue, greater > m;
ll max(ll a, ll b) { return a > b ? a : b; }
ll min(ll a, ll b) { return a < b ? a : b; }
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) { return a * b / gcd(a, b); }
map<ll, ll> mp;
ll ksc(ll a, ll b)
{
ll ans = 0;
while (b)
{
if (b & 1)
ans = (ans + a) % mod;
a = (a + a) % mod;
b >>= 1;
}
return ans;
}
ll ksm(ll a, ll b)
{
ll ans = 1ll;
while (b)
{
if (b & 1)
ans = ksc(ans, a) % mod;
a = ksc(a, a) % mod;
b >>= 1;
}
return ans;
}
ll dp[105][16005];
string p = "abacaba";
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
ll t;
cin >> t;
while (t--)
{
ll n;
cin >> n;
for (int i = 1; i <= n; i++)
{
cin >> a[i];
}
sort(a + 1, a + n + 1);
bool fla = true;
for (int i = 2; i <= n; i++)
{
if (abs(a[i] - a[i - 1]) > 1)
{
fla = false;
cout << "NO" << endl;
break;
}
}
if (fla)
cout << "YES" << endl;
}
}
B : G i f t s B:Gifts B:Gifts F i x i n g Fixing Fixing
思路：
求出两个数组的最小值，然后在每个位置上分别对每个数组的最小值做差，累加差的最大值。
参考代码：
#include p;
//priority_queue, greater > m;
ll max(ll a, ll b) { return a > b ? a : b; }
ll min(ll a, ll b) { return a < b ? a : b; }
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) { return a * b / gcd(a, b); }
map<ll, ll> mp;
ll ksc(ll a, ll b)
{
ll ans = 0;
while (b)
{
if (b & 1)
ans = (ans + a) % mod;
a = (a + a) % mod;
b >>= 1;
}
return ans;
}
ll ksm(ll a, ll b)
{
ll ans = 1ll;
while (b)
{
if (b & 1)
ans = ksc(ans, a) % mod;
a = ksc(a, a) % mod;
b >>= 1;
}
return ans;
}
ll dp[105][16005];
string p = "abacaba";
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
ll t;
cin >> t;
while (t--)
{
ll n, mina = 1e9 + 7, minb = 1e9 + 7;
cin >> n;
for (int i = 1; i <= n; i++)
cin >> a[i], mina = min(a[i], mina);
for (int i = 1; i <= n; i++)
cin >> b[i], minb = min(b[i], minb);
ll ans = 0;
for (int i = 1; i <= n; i++)
ans += max(a[i] - mina, b[i] - minb);
cout << ans << endl;
}
}
C : B o a t s C: Boats C:Boats C o m p e t i t i o n Competition Competition
思路：
首先得到所有可能两两相加的答案，然后进行去重处理，标记初始数组中每个元素出现的次数，并标记当前元素存在，然后暴力枚举处理后的两元素和-标记元素是否存在，根据数量求答案。
参考代码：
#include p;
//priority_queue, greater > m;
ll max(ll a, ll b) { return a > b ? a : b; }
ll min(ll a, ll b) { return a < b ? a : b; }
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) { return a * b / gcd(a, b); }
map<ll, ll> mp;
ll ksc(ll a, ll b)
{
ll ans = 0;
while (b)
{
if (b & 1)
ans = (ans + a) % mod;
a = (a + a) % mod;
b >>= 1;
}
return ans;
}
ll ksm(ll a, ll b)
{
ll ans = 1ll;
while (b)
{
if (b & 1)
ans = ksc(ans, a) % mod;
a = ksc(a, a) % mod;
b >>= 1;
}
return ans;
}
ll dp[105][16005];
string p = "abacaba";
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
ll t;
cin >> t;
while (t--)
{
ll n, ma = -1000, minb = 1e9 + 7, ans = 0;
cin >> n;
vec.clear();
memset(vis, 0, sizeof vis);
memset(b, 0, sizeof b);
for (ll i = 1; i <= n; i++)
cin >> a[i], vis[a[i]] = 1, b[a[i]]++;
sort(a + 1, a + 1 + n);
int cnt = 0;
for (int i = 1; i <= n; i++)
for (int j = i + 1; j <= n; j++)
vec.push_back(a[i] + a[j]);
sort(vec.begin(), vec.end());
vec.erase(unique(vec.begin(), vec.end()), vec.end());
int k = vec.size();
for (int i = 0; i < k; i++)
{
//cout << vec[i] << endl;
int t = 0;
for (int j = 1; j <= n; j++)
{
int kp = vec[i] - a[j];
if (kp > 0 && vis[kp])
{
if (kp == a[j] && (b[kp] - c[kp]) <= 1)
continue;
int x = min(b[a[j]] - c[a[j]], b[kp] - c[kp]);
//cout << x << "*****" << endl;
x = max(x, 0);
if (kp == a[j])
t += x / 2;
else
t += x;
c[a[j]] += x, c[kp] += x;
}
}
memset(c, 0, sizeof c);
ans = max(ans, t);
}
cout << ans << endl;
}
}
D : B i n a r y D:Binary D:Binary S t r i n g String String T o To To S u b s e q u e n c e s Subsequences Subsequences
思路：
队列+模拟
#include vec;
//typedef pair p;
//priority_queue, greater > m;
ll max(ll a, ll b) { return a > b ? a : b; }
ll min(ll a, ll b) { return a < b ? a : b; }
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) { return a * b / gcd(a, b); }
map<ll, ll> mp;
ll ksc(ll a, ll b)
{
ll vec = 0;
while (b)
{
if (b & 1)
vec = (vec + a) % mod;
a = (a + a) % mod;
b >>= 1;
}
return vec;
}
ll ksm(ll a, ll b)
{
ll vec = 1ll;
while (b)
{
if (b & 1)
vec = ksc(vec, a) % mod;
a = ksc(a, a) % mod;
b >>= 1;
}
return vec;
}
ll dp[105][16005];
string p = "abacaba";
queue<ll> qk, q;
vector<ll> vec;
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
ll t,n;
cin >> t;
string str;
while (t--)
{
cin >> n;
cin >> str;
vec.clear();
while (!q.empty())
q.pop();
while (!qk.empty())
qk.pop();
ll ans = 1;
for (int i = 0; i < n; i++)
{
if (str[i] == '0')
{
if (q.empty())
vec.push_back(ans++);
else
vec.push_back(vec[q.front()]), q.pop();
qk.push(i);
}
else
{
if (qk.empty())
vec.push_back(ans++);
else
vec.push_back(vec[qk.front()]), qk.pop();
q.push(i);
}
}
cout << ans - 1 << endl;
for (int i = 0; i < vec.size(); i++)
cout << vec[i] << ' ';
cout << endl;
}
}