Codeforces Round 907 (Div. 2)

Codeforces Round 907 (Div. 2)

B题注意到每次都会至少下降1，所以不会超过30次，直接O(30n)即可

C题感觉可能比D和F还要思维一些。
肯定是尽量多积累combo一些然后一次清空，那么我们能清空的最大值就是当前的最大值，所以每次用小的来累计combo，然后消除当前的最大值，即使小的那些不是一整段也没有关系，然后就是一些特判。复杂度O(n)

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<map>
#include<vector>
#include<set>
#define A puts("YES")
#define B puts("NO")
#define fo(i,a,b) for (int (i)=(a);(i)<=(b);(i)++)
#define fd(i,b,a) for (int (i)=(b);(i)>=(a);(i)--)
#define mk(x,y) make_pair((x),(y))
using namespace std;
typedef double db;
typedef long long ll;
const int mo=998244353;
//const int inf=1<<30;
const int N=2e5+5;
ll n,a[N],siz,ans;
int main()
{
//	freopen("data.in","r",stdin);
//	freopen("ans.out","w",stdout);
	
	int T;
	scanf("%d",&T);
	while (T--){
		scanf("%lld",&n);
		fo(i,1,n) scanf("%lld",&a[i]);
		sort(a+1,a+n+1);
		
		ans=0;
		int j=1;
		siz=n;
		ll s;
		while (1) {
			if (!a[j]) j++;
			if (j>n) break;
			
			if (j==n) {
				if (!a[n]) break;
				if (a[n]==1) ans++;
				else if (a[n]&1) ans+=a[n]/2+2;
				else ans+=a[n]/2+1;
				break;
			}
			s=a[j];
			
			while (j+1<n && s<a[n]) j++,s+=a[j];
			if (s<a[n]) {
				ans+=s;
				if (a[n]==s+1) ans+=2;
				else if ((a[n]+s)%2) ans+=(a[n]-s)/2+2;
				else ans+=(a[n]-s)/2+1;
				break;
			}
			
			ans+=a[n]+1;
			a[j]=(s-a[n]);
			n--;
		}
		
		printf("%lld\n",ans);
	}
	
	return 0;
}

 
 

D题感觉没什么好说的，直接两个log就完事了。

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<set>
#include<map>
#include<vector>
#define A puts("Yes")
#define B puts("No")
#define fo(i,a,b) for (ll (i)=(a);(i)<=(b);(i)++)
#define fd(i,b,a) for (int (i)=(b);(i)>=(a);(i)--)
#define mk(x,y) make_pair((x),(y))

using namespace std;
typedef double db;
typedef long long ll;
typedef __int128 i128;
const ll mo = 1e9 + 7;
const int N = 2e5 + 5;
// const ll inf = 2e18;
ll d, u, ans;
ll b[100];
void solve(ll x, ll d, ll u) {
    ll l, r, mid;
    ll now = 1, tot = 0;
    while ((i128)now * (i128)x <= u) now *= x, tot++;
    for (ll i = u;i <= u;i = l + 1) {
        l = i; r = d;

        while ((i128)now * (i128)x <= i) now *= x, tot++;
        while (l < r) {
            mid = (l + r + 1) >> 1;
            if (mid <= (i128)now * (i128)x) l = mid;
            else r = mid - 1;
        }
        ans += (l - i + 1) * tot;
    }
}
int main()
{
    // #ifdef LOCAL
    //  		freopen("in.txt", "r", stdin);
    //     	freopen("out.txt", "w", stdout);
    // #endif

    b[0] = 1;
    fo(i, 1, 60) b[i] = b[i - 1] * 2ll;

    int T;
    scanf("%d", &T);
    while (T--) {
        ans = 0;
        scanf("%lld %lld", &d, &u);
        fo(i, 2, 60) {
            if (b[i] > u) break;
            // if (b[i] < d) continue;
            solve(i, max(d, b[i]), min(u, b[i + 1] - 1));
        }

        printf("%lld\n", ans);
    }

    return 0;
}

F是个非常经典的套路，假如没有时间的限制，那么就是直接利用用差分+dfs序套个fenwick树就行，加上时间限制也简单，直接将询问倒着做就行，在一个点被加入的时候计算它的答案。

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<map>
#include<vector>
#include<set>
#define A puts("YES")
#define B puts("NO")
#define fo(i,a,b) for (int (i)=(a);(i)<=(b);(i)++)
#define fd(i,b,a) for (int (i)=(b);(i)>=(a);(i)--)
#define mk(x,y) make_pair((x),(y))
using namespace std;
typedef double db;
typedef long long ll;
const int mo=998244353;
//const int inf=1<<30;
const int N=5e5+5;
struct node{
	int t,x,v,y;
};
node q[N];
int n,m;
int head[N],to[N],nex[N],tot;
int l[N],r[N],cnt;
ll c[N],y,x,ans[N];

int lowbit(int x){
	return x&(-x);
}
void upd(int x,ll y){
	for (;x<N;x+=lowbit(x)) c[x]+=y;
}
ll ask(int x){
	ll s=0;
	for (;x;x-=lowbit(x)) s+=c[x];
	return s;
}
void add(int x,int y){
	to[++tot]=y; nex[tot]=head[x]; head[x]=tot;
}
void dfs(int x){
	ans[x]=0;
	l[x]=r[x]=++cnt;
	for (int i=head[x];i;i=nex[i]){
		int v=to[i];
		dfs(v);
		r[x]=max(r[x],r[v]);
	}
}
int main()
{
//	freopen("data.in","r",stdin);
//	freopen("ans.out","w",stdout);

	int T;
	scanf("%d",&T);
	while (T--){
		scanf("%d",&m);
		
		n=1;
		head[1]=0; 
		tot=0;
		
		fo(i,1,m) {
			scanf("%d",&q[i].t);
			if (q[i].t==1) {
				scanf("%d",&q[i].x);
				n++;
				add(q[i].x, n);
				q[i].y=n;
				head[n]=0;
			}
			else {
//				R(q[i].x); R(q[i].v);
				scanf("%d %d",&q[i].x, &q[i].v);
			}
		}
		
		cnt=0;
		dfs(1);
		fo(i,1,n+3) c[i]=0;

		fd(i,m,1) {
			if (q[i].t==1) {
				ans[q[i].y]=ask(l[q[i].y]);
			}
			else {
				x=q[i].x; y=q[i].v;
				upd(l[x], y);
				upd(r[x]+1, -y);
			}
		}
		
		ans[1]=ask(1);
		fo(i,1,n) printf("%lld ",ans[i]);
		printf("\n");

	}
	return 0;
}