第一题用搜索,超时了,待补
##7-1搜索 12/20分
#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\n",ans[i].address,ans[i].data,ans[i+1].address);
}
printf("%05d %d -1\n",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;
}
原文地址:https://www.cnblogs.com/fisherss/p/11615671.html
时间: 2024-10-07 12:34:51