PAT（甲级）2019年秋季考试

第一题用搜索，超时了，待补

更新第一题思路
dfs + 剪枝，首先确定 n的最后一位数字肯定是9，为什么呢，因为 任意两个相邻的数肯定互为质数(gcd=1)，所以 n 的末尾肯定是9，这样n+1产生的各个位数和相加的和 才能有可能和sum(n)产生gcd>2的素数，
所以k可以剪枝成k-1，缩小一位，复杂度降低很多。。。这样估计就能AC了。
另外判断n和n+1两个数的各个位数之和的gcd是否为大于2的素数，可以提前预处理(数据量<90)，
第一题参考链接

7-1搜索 12/20分

原来写的代码，超时2个测试点

#include<bits/stdc++.h>
using namespace std;

int n,k,m;
typedef long long ll;
ll starts = 0;
ll endss = 0;
bool flag = false;
bool first = true;
int t = 1;

int sums(ll x){
	int ans = 0;
	while(x){
		ans += x%10;
		x/=10;
	}
	return ans;
}


ll gcd(ll a,ll b){
	if(b == 0) return a;
	return gcd(b,a%b);
}

bool solve(ll x){
	if(x <= 2) return false;
	for(int i=2;i<=sqrt(x);i++){
		if(x%i == 0) return false;
	}
	return true;
}


void dfs(int step,ll x){
	if(x >= endss) return;
	if(step > k+1) return;
	if(sums(x) > m) return;
	if(step == k+1) {
		int t1 = sums(x);
		if(sums(x) != m) return;
		int t2 = sums(x+1);
		ll g = gcd(t1,t2);
		if(solve(g)){
			if(first){
				first = false;
				cout<<"Case "<<t<<endl;
			}
			flag = true;
			cout<<t2<<" "<<x<<endl;
		}
		return;
	}
	for(int i=0;i<=9;i++){
		if(i==0 && step == 1) continue;
		ll temp = x * 10 + i;
		dfs(step+1,temp);
	}
} 

int main(){
	cin>>n;
	for(int i=1;i<=n;i++){
		cin>>k>>m;
		starts = pow(10,k-1);
		endss = pow(10,k);
		flag = false;	
		first = true;
		t = i;
		dfs(1,0);
		if(flag == false){
			cout<<"Case "<<t<<endl;
			cout<<"No Solution"<<endl;
		}
	}
	return 0;
}

7-2链表模拟 25

#include<bits/stdc++.h>
using namespace std;

const int maxn = 1e5+10;

struct node{
	int  address;
	int data;
	int  next;
}li[maxn];

int s1,s2;
int n; //不一定都是有效结点 
vector<node> v1;
vector<node> v2;
vector<node> ans;

int main(){
	scanf("%d%d%d",&s1,&s2,&n);
	for(int i=1;i<=n;i++){
		int dat,add,nex;
		scanf("%d%d%d",&add,&dat,&nex);
		li[add].address = add;
		li[add].data = dat;
		li[add].next = nex;
	}
	for(int cur = s1;cur != -1;cur = li[cur].next) v1.push_back(li[cur]);
	for(int cur = s2;cur != -1;cur = li[cur].next) v2.push_back(li[cur]);
	int len1 = v1.size() - 1;
	int len2 = v2.size() - 1;
	if(len1 >= 2*len2){
		reverse(v2.begin(),v2.end());
		int idx2 = 0;
		int idx1 = 0;
		while(idx2 <= len2){
			ans.push_back(v1[idx1++]);
			ans.push_back(v1[idx1++]);
			ans.push_back(v2[idx2++]);
		}
		while(idx1 <= len1) ans.push_back(v1[idx1++]);
	}else{
		reverse(v1.begin(),v1.end());
		int idx2 = 0;
		int idx1 = 0;
		while(idx1 <= len1){
			ans.push_back(v2[idx2++]);
			ans.push_back(v2[idx2++]);
			ans.push_back(v1[idx1++]);
		}
		while(idx2 <= len2) ans.push_back(v2[idx2++]);
	}
	n = ans.size();
	for(int i=0;i<n-1;i++){
		printf("%05d %d %05d
",ans[i].address,ans[i].data,ans[i+1].address);
	}
	printf("%05d %d -1
",ans[n-1].address,ans[n-1].data);
	return 0;
}

7-3树 25

#include<bits/stdc++.h>
using namespace std;

struct node{
	string dat;
	int lc,rc;
}tree[25];

int n;
int vis[25];

void dfs(int x){
	if(x > n) return;
	cout<<"(";
	if(tree[x].lc == -1 &&  tree[x].rc != -1){
		cout<<tree[x].dat;
		dfs(tree[x].rc);
		cout<<")";
	}else{
		if(tree[x].lc!=-1) dfs(tree[x].lc);
		if(tree[x].rc!=-1) dfs(tree[x].rc);
		cout<<tree[x].dat;
		cout<<")";
	}
}

int main(){
	cin>>n;
	for(int i=1;i<=n;i++){
		cin>>tree[i].dat>>tree[i].lc>>tree[i].rc;
		if(tree[i].lc != -1) vis[tree[i].lc] = 1;
		if(tree[i].rc != -1) vis[tree[i].rc] = 1;
	}
	int root = 1;
	for(int i=1;i<=n;i++){ //先确定 根的结点 没有当过 
		if(!vis[i]) root = i;
	}
	dfs(root);
	return 0;
}

7-4最短路 30

#include<bits/stdc++.h>
using namespace std;

const int inf = 0x3f3f3f3f;
const int maxn = 1e3+10;
const int maxm = 1e5+10;
struct edge{
	int v;
	int w;
	edge(int vv,int ww){
		v = vv;
		w = ww;
	}
};

vector<edge> g[maxn];
int n,e,k;
int di[maxn];
int dist[maxn];
int vis[maxn];

bool dijkstra(int s){
	memset(dist,inf,sizeof(dist));
	memset(vis,0,sizeof(vis));
	dist[s] = 0;
	for(int i=1;i<=n;i++){
		int v,min_w = inf;
		for(int j=1;j<=n;j++){
			if(!vis[j] && dist[j] < min_w){
				v = j;
				min_w = dist[j];
			}
		}
		if(min_w == inf) return false;
		vis[v] = 1;
		for(int j=0;j<g[v].size();j++){
			int x = g[v][j].v;
			int w = g[v][j].w;
			if(!vis[x] && dist[x] >dist[v] + w){
				dist[x] = dist[v] + w;
			}
		}
	}
	return true;
}
int vis2[maxn];
bool solve(){
	memset(vis2,0,sizeof(vis2));
	vis2[di[1]] = 1;
	for(int i=2;i<=n;i++){
		int mind = inf;
		for(int j=1;j<=n;j++){
			if(!vis2[j] && dist[j] <= mind){
				mind = dist[j];
			}
		}
		if(mind != dist[di[i]]) 
			return false;
		vis2[di[i]] = 1;
	}
	return true;
}


void solve2(){
	int visit[n + 1];
	for (int i = 1; i <= n; i++) visit[i] = 0;
	int start = di[1];
	visit[start] = 1;
	bool flag = false;
	for (int i = 2; i <= n; i++) {
		int mine = dist[di[i]];
		visit[di[i]] = 1;
		for (int j = 1; j <= n; j++) {
			if (visit[di[j]]) continue;
			int curv = dist[di[j]];
			if (mine >= curv) mine = curv;
		}
		if (mine < dist[di[i]]) {
			cout << "No" << endl;
			flag = true;
			break;
		}
	}
	if (flag == false) cout << "Yes" << endl;
}

int main(){
	cin>>n>>e;
	for(int i=1;i<=e;i++){
		int u,v,w;
		cin>>u>>v>>w;
		g[u].push_back(edge(v,w));
		g[v].push_back(edge(u,w));
	}
	cin>>k;
	for(int i=1;i<=k;i++){
		for(int j=1;j<=n;j++) cin>>di[j];
		int st = di[1];
		dijkstra(st);
//		solve2();
		if(solve()){
			cout<<"Yes"<<endl;
		}else{
			cout<<"No"<<endl;
		}
	}
	return 0;
}