Nowcoder 提高 Day1

比赛链接

A 中位数(前缀和 二分)

额，确实没想到逼近...
然后写了n^2log的暴力,还CE了
只需要判断是否能有大于当前mid的中位数就好
这显然是可以二分的

代码

#include<bits/stdc++.h>
using namespace std;
#define li long long
#define gc getchar()
#define pc putchar
li read(){
    li x = 0,y = 0,c = gc;
    while(!isdigit(c)) y = c,c = gc;
    while(isdigit(c)) x = (x << 1) + (x << 3) + (c ^ '0'),c = gc;
    return y == '-' ? -x : x;
}
void print(li q){
    if(q < 0){
        pc('-');
        q = -q;
    }
    if(q >= 10) print(q / 10);
    pc(q % 10 + '0');
}
int n,m;
int a[100010],b[100010],c[100010],d[100010];
bool chk(int p){
    int i,j;
    for(i = 1;i <= n;++i) b[i] = a[i] >= p ? 1 : -1;
    for(i = 1;i <= n;++i) c[i] = c[i - 1] + b[i];
    for(i = 1;i <= n;++i) d[i] = min(d[i - 1],c[i]);
    for(i = n,j = -n;i >= m;--i){
        j = max(j,c[i]);
        if(j - d[i - m] > 0) return 1;
    }
    return 0;
}
int main(){
    int i,j;
    n = read();m = read();
    for(i = 1;i <= n;++i) a[i] = read();
    int l = 1,r = 1000000000,mid,as = 1;
    while(l <= r){
        mid = l + r >> 1;
        if(chk(mid)) as = mid,l = mid + 1;
        else r = mid - 1;
    }
    print(as);
    return 0;
}

B 数数字(数位DP)

数位DP不会
不过yy一下，写个暴力dp,map转移似乎就...过了

代码

#include<iostream>
#include<cstdio>
#include<map>
#define LL long long
using namespace std;
inline LL read() {
    char c = getchar(); LL x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
LL L, R, L1, R1;
map<LL, LL> f[20];  
LL s[21], num = 0;
LL dfs(int x, int lim, LL mul) {
    if(x < 0) return 0;
    if(x == 0) {
        if(mul == -1) mul = 0;
        return mul >= L1 && mul <= R1;
    }
    if((!lim) && (f[x].find(mul) != f[x].end())) return f[x][mul];
    LL ans = 0;
    for(int i = 0; i <= (lim ? s[x] : 9); i++) {
        if(mul == -1) {
            if(i == 0) ans += dfs(x - 1, 0, -1);//tag
            else ans += dfs(x - 1, lim && (i == s[x]), i);
        } else ans += dfs(x - 1, lim && (i == s[x]), i * mul);
    }
    if(!lim) f[x][mul] = ans;
    return ans;
}
LL solve(LL x) {
    if(x == -1) return 0;
    num = 0;
    while(x) s[++num] = x % 10, x /= 10;
    return dfs(num, 1, -1);
}
int main() {
    L = read(); R = read(); L1 = read(); R1 = read();
    LL ans = solve(R) - solve(L - 1);
    cout << ans;
    return 0;
}

C 保护(dfs序 主席树)

额，主席树维护一下子树中点能覆盖的第k小深度..就没了

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn=200005;

struct sj{int to,next;}line[maxn*2];
int size,n,m,num,q,siz[maxn];
int head[maxn],id[maxn],tot,idx[maxn];
int a[maxn],b[maxn],c[maxn],T[maxn];
int L[maxn * 64],R[maxn * 64],sum[maxn * 64];
void add(int x,int y) { 
    line[++size].to=y;
    line[size].next=head[x];
    head[x]=size;
}
int old[maxn],deep[maxn]; 
int dad[maxn][27]; 
void dfs(int x,int fa = 0) { 
    siz[x] = 1;
	++ num;  
    id[x]=num; 
    old[num]=x;
    deep[x] = deep[fa] + 1; 
    for(int i=head[x];i;i=line[i].next) {
        int tt=line[i].to;
        if(!siz[tt]) {dad[tt][0] = x; 
            dfs(tt,x);
            siz[x]+=siz[tt];
		}  
    } 
} 
inline int lca(int x,int y)  {
	if(deep[x] > deep[y]) std::swap(x,y); 
	for(int i = 20;i >= 0;-- i) if(deep[dad[y][i]] >= deep[x]) y = dad[y][i]; 
	if(x == y) return x; 
	for(int i = 20;i >= 0;-- i) 
		if(dad[x][i] != dad[y][i]) 
			x = dad[x][i],y = dad[y][i]; 
	return dad[x][0]; 
} 
int build(int l,int r) {
    int rt=++tot;
    if(l!=r) {
        int mid=(l+r)/2;
        L[rt]=build(l,mid);
        R[rt]=build(mid+1,r);
    }
    return rt;
}

int update(int pre,int l,int r,int x) { 
    int rt =++ tot;
    L[rt]=L[pre]; 
    R[rt]=R[pre];
    sum[rt]=sum[pre] + 1; 
    int mid=(l+r)/2;
    if(l!=r) {
        if (x<=mid) L[rt]=update(L[pre],l,mid,x);
        else R[rt]=update(R[pre],mid+1,r,x);
    }
    return rt;
}

int query(int x,int y,int l,int r,int k) {
    if(l==r)return l;
    int now=sum[L[y]]-sum[L[x]];
    int mid=(l+r)/2;
    if(now>=k) return query(L[x],L[y],l,mid,k);
    else return query(R[x],R[y],mid+1,r,k-now);
}
vector<int> qq[maxn]; 
int main()  { 
    scanf("%d%d",&n,&m); 
    for(int i=1;i<n;i++) { 
        int x,y; scanf("%d%d",&x,&y); 
        add(x,y); add(y,x); 
    }
    dfs(1); 
    for(int i = 1;i <= 20;++ i) 
		for(int x = 1;x <= n;++ x) dad[x][i] = dad[dad[x][i - 1]][i - 1]; 
    for(int u,v,i = 1;i <= m;++ i) { 
		scanf("%d%d",&u,&v); 
		int l = lca(u,v); 
		qq[u].push_back(deep[l]); 
		qq[v].push_back(deep[l]); 
	} 
    T[0]=build(1,n + 1); 
    for(int i=1;i <= n;i ++)  { 
    	if(qq[old[i]].size()) { 
    		for(int j = 0;j < qq[old[i]].size();++ j) { 
        		if(j == 0) T[i] = update(T[i - 1],1,n + 1,qq[old[i]][j]); 
				else T[i] = update(T[i],1,n + 1,qq[old[i]][j]); 
			} 
    	} 
    	else T[i] = update(T[i - 1],1,n + 1,n + 1); 
	} 
    scanf("%d",&q); 
    while(q--)  { 
        int x,k; 
        scanf("%d%d",&x,&k);
        int p=query(T[id[x]-1],T[id[x]+siz[x] - 1],1,n + 1,k);
        printf("%d
",std::max(deep[x] - p,0)); 
    }           
    return 0; 
}