A. Be Positive
code
#include<bits/stdc++.h>
using namespace std;
const int NN = 1e4;
int T;
int n,a[NN];
int cnt1,cnt2;
void solve(){cnt1 = cnt2 = 0;cin >> n;for(int i = 1; i <= n; ++i){cin >> a[i];if(a[i] == -1) ++cnt1;else if(a[i] == 0) ++cnt2;}cout << cnt2 + (cnt1&1) * 2 << '\n';
}
int main(){cin >> T;while(T--){solve();}
}
B. Unconventional Pairs
code
#include<bits/stdc++.h>
using namespace std;
const int NN = 2e5 + 8;
int t,n;
int a[NN];
void solve(){cin >> n;for(int i = 1; i <= n; ++i){cin >> a[i];}sort(a+1,a+1+n);int ans = 0;for(int i = 1; i <= n/2; ++i){ans = max(a[i*2]-a[i*2-1],ans);}cout << ans << '\n';
}
int main(){ios::sync_with_stdio(false),cin.tie(0);cin >> t;while(t--){solve();}
}
C. MEX rose
可以乱改求 \(mex\)
code
#include<bits/stdc++.h>
using namespace std;
const int NN = 2e5 + 8;
int t;
int n,k;
int a[NN];
void solve(){int cnt1 = 0,cnt2 = 0;cin >> n >> k;for(int i = 1; i <= n; ++i){cin >> a[i];if(a[i] == k) ++cnt1;}sort(a+1,a+1+n);for(int i = 1; i <= n; ++i){if(a[i] < k && a[i] != a[i-1]) ++cnt2;}cout << max(cnt1,k-cnt2) << '\n';
}
int main(){ios::sync_with_stdio(false),cin.tie(0);cin >> t;a[0] = -1;while(t--){solve();}
}
D. A and B
求让一种字符连在一起的最小步数
我们首先提取出 \(a,b\) 两种字符的下标组，并且重新编辑位置为该字母前面与它不同的字符数量
然后我们最后的问题就变为了每一次操作 \(+1/-1\) 求最小的使得所有数相同的操作次数。
这个问题十分经典，就是让所有数变成中位数即可
code
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int NN = 2e5 + 8;
int t,n;
int a[NN],b[NN],lena,lenb;
string s;
void solve(){cin >> n; cin >> s;lena = lenb = 0;for(int i = 0,dela = 0,delb = 0; i < n; ++i){if(s[i] == 'a'){a[++lena] = i-dela;++dela;}else{b[++lenb] = i-delb;++delb;}}int mda = (lena + 1) / 2,mdb = (lenb + 1) / 2;ll ansa = 0,ansb = 0;for(int i = 1; i <= lena; ++i){ansa += abs(a[i]-a[mda]);}for(int i = 1; i <= lenb; ++i){ansb += abs(b[i]-b[mdb]);}cout << min(ansa,ansb) << '\n';
}
int main(){ios::sync_with_stdio(false),cin.tie(0);cin >> t;while(t--){solve();}
}
E. Hidden Knowledge of the Ancients
求长度为 \([l,r]\) 的有 \(k\) 种数的子区间的个数
第一种方法是两套双指针，然后对于每一个 \(l\)，维护从 \(l\) 开始有 \(k\) 种数的最小区间和最大区间
本人写了第二种复杂很多的算法，我们可以发现，一个对于一个区间元素种类能够做出贡献，当且仅当在这个包含它的区间内并且在它前面没有和它相同的数（相当于一个矩形加）
然后我们需要找到有 \(k\) 种元素的区间个数，这个可以通过离线询问 + 扫描线解决
code
#include<bits/stdc++.h>
using namespace std;
const int NN = 2e5 + 8;
int t;
struct Seg{int l,r;int maxn,minn,tag;#define ls(x) (x << 1)#define rs(x) (x << 1 | 1)#define l(x) tree[x].l#define r(x) tree[x].r#define maxn(x) tree[x].maxn#define minn(x) tree[x].minn#define tag(x) tree[x].tag
}tree[NN << 2];
void addlz(int x,int num){maxn(x) += num; minn(x) += num;tag(x) += num;
}
void pushdown(int x){addlz(ls(x),tag(x)), addlz(rs(x),tag(x));tag(x) = 0;
}
void pushup(int x){maxn(x) = max(maxn(ls(x)),maxn(rs(x)));minn(x) = min(minn(ls(x)),minn(rs(x)));
}
void build(int x,int l,int r){l(x) = l;r(x) = r;minn(x) = maxn(x) = tag(x) = 0;if(l == r) return;int mid = (l + r) / 2;build(ls(x),l,mid);build(rs(x),mid+1,r);return;
}
void modify(int x,int l,int r,int num){if(l <= l(x) && r(x) <= r){addlz(x,num);return;}int mid = (l(x) + r(x)) / 2;pushdown(x);if(l <= mid) modify(ls(x),l,r,num);if(mid+1 <= r) modify(rs(x),l,r,num);pushup(x);
}
int findl(int x,int num){if(l(x) == r(x)) return maxn(x) == num ? l(x) : -1;pushdown(x);if(maxn(ls(x)) >= num) return findl(ls(x),num);else return findl(rs(x),num);
}
int findr(int x,int num){if(l(x) == r(x)) return minn(x) == num ? l(x) : -1;pushdown(x);if(minn(rs(x)) <= num) return findr(rs(x),num);else return findr(ls(x),num);
}int n,k,l,r;
int a[NN];
int pre[NN],b[NN],lenb;
struct Oper{int pos,num;
};
vector<Oper> op[NN];
void solve(){long long ans = 0;cin >> n >> k >> l >> r;build(1,1,n);op[0].clear();for(int i = 1; i <= n; ++i){cin >> a[i];b[i] = a[i];pre[i] = 0;op[i].clear();}sort(b+1,b+1+n);lenb = unique(b+1,b+1+n) - b - 1;for(int i = 1; i <= n; ++i) a[i] = lower_bound(b+1,b+1+lenb,a[i]) - b;for(int i = 1; i <= n; ++i){op[pre[a[i]]].push_back({i,1});op[i].push_back({i,-1});pre[a[i]] = i;}for(int i = 0; i <= n; ++i){int fr = findr(1,k),fl = findl(1,k);if(i != 0 && fr != -1 && fl != -1){fr = min(i+r-1,fr);fl = max(i+l-1,fl);ans += max(0,fr - fl + 1);}for(int j = 0; j < op[i].size(); ++j){modify(1,op[i][j].pos,n,op[i][j].num);}}cout << ans << '\n';
}int main(){ios::sync_with_stdio(false),cin.tie(0);cin >> t;while(t--){solve();}
}
F. Nezuko in the Clearing
tag:二分答案
我们首先可以贪心地发现，如果说我们知道最后的休息次数，那么我们只需要让每次的连续走的步数更加平均即可，这样一定就可以让花费的生命值最小
那么我们知道了验证方法，我们进行二分答案即可
code
#include<bits/stdc++.h>
using namespace std;
#define int long long
const int NN = 0;
int t,h,d;
bool check(int num){//二分休息次数 int stlen = d / (num+1), add = d % (num+1);return num + h - (num + 1) * (stlen * (stlen + 1) / 2) - add * (stlen + 1) >= 0;
}
void solve(){cin >> h >> d;int st = 0;if(h == 1) st = 1;else --h;int ans = 0,l = 0,r = d;while(l <= r){int mid = (l + r) / 2;if(check(mid)) ans = mid,r = mid - 1;else l = mid + 1;}cout << ans + st + d << '\n';
}
signed main(){ios::sync_with_stdio(false),cin.tie(0);cin >> t;while(t--){solve();}
}
