11.17 解题报告

T1

用时：1 h
期望得分：\(60\) pts
实际得分：\(30\) pts

要求统计数组 \(a\) 中有序三元组 \((x,y,z)\) 的个数，满足 \(\gcd(a_x,a_y)=a_z\)，直接枚举 \(x\)，\(y\)，将 \(x\) 后面的加入一个 map 中，统计答案即可。

#include<bits/stdc++.h>
#define ll long long
//#define int long long
//#define ull unsigned long long
#define lc(k) k<<1
#define rc(k) k<<1|1
#define lb lower_bound
#define orz cout<<"gzn ak ioi\n"
const int MAX=1e6+10;
const int MOD=1e9+7;
using namespace std;
inline char readchar() {
	static char buf[100000], *p1 = buf, *p2 = buf;
	return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF : *p1++;
}
inline int read() {
//#define readchar getchar
	int res = 0, f = 0;
	char ch = readchar();
	for(; !isdigit(ch); ch = readchar()) if(ch == '-') f = 1;
	for(; isdigit(ch);ch = readchar()) res = (res << 1) + (res << 3) + (ch ^ '0');
	return f ? -res : res;
}
inline void write(int x) {
    if(x<0){putchar('-');x=-x;}
    if(x>9) write(x/10);
    putchar(x%10+'0');
}
map<int,int> mp;
int a[3010];
signed main(){
//	freopen("1.in","r",stdin);
//	freopen("mobius.in","r",stdin);
//	freopen("mobius.out","w",stdout);
	int n=read();
	for(register int i=1;i<=n;i++) a[i]=read();
	ll sum=0;
	for(register int i=1;i<=n;i++){
		mp.clear();
		for(register int j=i+1;j<=n;j++) mp[a[j]]++;
		for(register int j=i+1;j<=n;j++){
			mp[a[j]]--;
			sum+=(mp[__gcd(a[i],a[j])]);
		}
	}
	cout<<sum;
	return 0;
}
/*

5
6 4 9 2 3

*/

T2

用时：\(2\) h

当 \(k\) 能够满足时，\(k+1\) 一定也能够满足，所以具有单调性，考虑二分，check时一个dp统计最长合法子序列长度是否 \(\ge m\) 即可。

#include<bits/stdc++.h>
#define ll long long
//#define int long long
//#define ull unsigned long long
#define lc(k) k<<1
#define rc(k) k<<1|1
#define lb lower_bound
#define orz cout<<"gzn ak ioi\n"
const int MAX=1e6+10;
const int MOD=1e9+7;
using namespace std;
inline char readchar() {
	static char buf[100000], *p1 = buf, *p2 = buf;
	return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF : *p1++;
}
inline int read() {
#define readchar getchar
	int res = 0, f = 0;
	char ch = readchar();
	for(; !isdigit(ch); ch = readchar()) if(ch == '-') f = 1;
	for(; isdigit(ch);ch = readchar()) res = (res << 1) + (res << 3) + (ch ^ '0');
	return f ? -res : res;
}
inline void write(int x) {
    if(x<0){putchar('-');x=-x;}
    if(x>9) write(x/10);
    putchar(x%10+'0');
}
/*
显然若k满足，k+1一定满足，考虑二分这个k
dp[i]表示以i结尾，符合条件的最长长度
dp[i]=max(dp[j])+1 1<=j<i |a[j]-a[i]|<=k
期望100pts 
*/
int n,m,a[1010],dp[1010];
bool check(int k){
	int ret=1;
	for(int i=1;i<=n;i++){
		dp[i]=1;
		for(int j=1;j<i;j++)
			if(abs(a[j]-a[i])<=k)
				dp[i]=max(dp[i],dp[j]+1);
		ret=max(ret,dp[i]);
	}
	return ret>=m;
}
signed main(){
//	freopen("cauchy.in","r",stdin);
//	freopen("cauchy.out","w",stdout);
	int l=1,r=2e9,ans=0;
	n=read(),m=read();
	for(int i=1;i<=n;i++) a[i]=read();
	while(l<=r){
		int mid=(l+r)>>1;
		if(check(mid)){
			ans=mid;
			r=mid-1;
		}
		else l=mid+1;
	}
	cout<<ans;
	return 0;
}

T3

用时：\(1.5\) h

发现由于题目的限制，必须要在一棵子树全部处理完成之后再去处理别的子树，所以这可以 dp。

这题用的时间比较久，在考场上想到一点思路，当时也考虑到了应该贪心的按 dp 值或者是剩余石子数来排序，但是没想打其实是按照 \(dp-w\) 来排序，然后 dp，由于细节太多，打了一个暴力就跑路了。

#include<bits/stdc++.h>
#define ll long long
#define int long long
//#define ull unsigned long long
#define lc(k) k<<1
#define rc(k) k<<1|1
#define lb lower_bound
#define orz cout<<"gzn ak ioi\n"
const int MAX=1e6+10;
const int MOD=1e9+7;
using namespace std;
inline char readchar() {
	static char buf[100000], *p1 = buf, *p2 = buf;
	return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF : *p1++;
}
inline int read() {
#define readchar getchar
	int res = 0, f = 0;
	char ch = readchar();
	for(; !isdigit(ch); ch = readchar()) if(ch == '-') f = 1;
	for(; isdigit(ch);ch = readchar()) res = (res << 1) + (res << 3) + (ch ^ '0');
	return f ? -res : res;
}
inline void write(int x) {
    if(x<0){putchar('-');x=-x;}
    if(x>9) write(x/10);
    putchar(x%10+'0');
}
int dp[MAX],n,head[MAX],cnt,w[MAX];
struct node{int u,v,net;}e[MAX];
void add(int u,int v){
	e[++cnt]=(node){u,v,head[u]};
	head[u]=cnt;
	return ;
}
pair<int,int> a[MAX];
void dfs(int u){
	int tp=0;
	for(int i=head[u];i;i=e[i].net){
		int v=e[i].v;
		dfs(v);
//		a[++tp]=make_pair(dp[v]-w[v],v);
	}
	for(int i=head[u];i;i=e[i].net){
		int v=e[i].v;
//		dfs(v);
		a[++tp]=make_pair(dp[v]-w[v],v);
	}
	sort(a+1,a+1+tp);
	int res=0;
//	cout<<u<<endl; 
	for(int i=tp;i>=1;i--){
		int v=a[i].second;
//		cout<<v<<" ";
		if(res>=dp[v]) res-=w[v];
		else{
			dp[u]+=dp[v]-res;
			res=dp[v]-w[v];
		}
	}
//	puts("\n--------------------");
	dp[u]+=max(0ll,w[u]-res);
	return ;
}
signed main(){
	n=read();
	for(int i=2;i<=n;i++){
		int x=read();
		add(x,i);
	}
	for(int i=1;i<=n;i++) w[i]=read();
	dfs(1);
	for(int i=1;i<=n;i++) cout<<dp[i]<<" ";
	return 0;
}