2024 蓝桥杯模拟赛3（div1+div2）
A.
暴力枚举
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N=2e5+10;
#define inf 0x3f3f3f3f
void solve() {
int n,k;cin>>n>>k;
vector<int>a(n+1);
for(int i=1;i<=n;i++)cin>>a[i];
int ans=0;
for(int i=1;i<n;i++){
for(int j=i+1;j<=n;j++){
if(a[i]*a[j]<=k)ans++;
}
}
cout<<ans;
}
signed main(){
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
int left=1;
//cin>>left;
while(left--){
solve();
}
}
B.
先按一周的算，剩下不足一周的再暴力
注意数据范围很大，用int128
#include <bits/stdc++.h>
using namespace std;
#define int __int128
const int N=2e5+10;
#define inf 0x3f3f3f3f
inline void read(__int128 &X)
{
X = 0;
int w=0; char ch=0;
while(!isdigit(ch)) {w|=ch=='-';ch=getchar();}
while(isdigit(ch)) X=(X<<3)+(X<<1)+(ch^48),ch=getchar();
if (w) X = -X;
}
void print(__int128 x)
{
if (!x) return ;
if (x < 0) putchar('-'),x = -x;
print(x / 10);
putchar(x % 10 + '0');
}
void solve() {
int a,b,n;read(a);read(b);read(n);
int tmp=a*5+b*2;
int l=1,r=1e18,mid;
int t=0;
// cout<<a<<' '<<b<<' '<<n<<"&&&"<<'\n';
while(l<=r){
mid=(l+r)/2;
if(mid*tmp<=n){
l=mid+1;
t=mid;
// cout<<"mid*tmp="<<mid*tmp<<'\n';
}
else r=mid-1;
}
// cout<<"t="<<t<<'\n';
// cout<<"t*tmp="<<t*tmp<<'\n';
if(t*tmp==n){
t*=7;
print(t);
return ;
}
int ans=t*7;
// cout<<"n="<<n<<'\n';
n-=t*tmp;
//cout<<"n="<<n<<'\n';
int tt=0;
for(int i=1;i<=7;i++){
if(i>=1&&i<=5){
tt+=a;
// cout<<"n="<<n<<'\n';
if(tt>=n){
ans+=i;
print(ans);
return ;
}
}else{
tt+=b;
if(tt>=n){
ans+=i;
print(ans);
return ;
}
}
}
}
signed main(){
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
int left=1;
//cin>>left;
while(left--){
solve();
}
}
C.
分奇数行和偶数行
奇数行是正常排列，偶数行是反着排
先算出m和n的位置再求横竖距离差
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N=2e5+10;
#define inf 0x3f3f3f3f
void solve() {
int w,m,n;cin>>w>>m>>n;
int up,down,l,r;
if(m%w==0)up=m/w;
else up=m/w+1;
if(n%w==0)down=n/w;
else down=n/w+1;
if(up&1)l=m-(up-1)*w;
else l=w-(m-(up-1)*w)+1;
if(down&1)r=n-(down-1)*w;
else r=w-(n-(down-1)*w)+1;
cout<<(abs(up-down)+abs(l-r));
}
signed main(){
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
int left=1;
//cin>>left;
while(left--){
solve();
}
}
D.
可以先倒退求0年的天干地支年，这样好处理许多
然后年份模上60，再暴力
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N=2e5+10;
#define inf 0x3f3f3f3f
vector<string>tg,dz;
void solve() {
tg.push_back("jia");tg.push_back("yi");tg.push_back("bing");tg.push_back("ding");
tg.push_back("wu");tg.push_back("ji");tg.push_back("geng");tg.push_back("xin");
tg.push_back("ren");tg.push_back("gui");
dz.push_back("zi");dz.push_back("chou");dz.push_back("yin");dz.push_back("mao");
dz.push_back("chen");dz.push_back("si");dz.push_back("wu");dz.push_back("wei");
dz.push_back("shen");dz.push_back("you");dz.push_back("xu");dz.push_back("hai");
int n;cin>>n;
n%=60;
for(int i=6,j=0;j<=n;i++,j++){
if(j==n) {
int ii = i % 10;
cout<<tg[ii];
break;
}
}
for(int i=8,j=0;j<=n;i++,j++){
if(j==n) {
int ii = i % 12;
cout<<dz[ii];
break;
}
}
}
signed main(){
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
int left=1;
//cin>>left;
while(left--){
solve();
}
}
E.
三层暴力循环，最后一层可以直接判
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N=2e5+10;
#define inf 0x3f3f3f3f
void solve() {
int n;cin>>n;
for(int i=0;i*i<=n;i++){
for(int j=0;i*i+j*j<=n;j++){
for(int k=0;i*i+j*j+k*k<=n;k++){
int tmp=sqrt(n-i*i-j*j-k*k);
if(tmp>=0){
if(tmp*tmp==n-i*i-j*j-k*k){
cout<<i<<' '<<j<<' '<<k<<' '<<tmp;
return ;
}
}
}
}
}
}
signed main(){
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
int left=1;
//cin>>left;
while(left--){
solve();
}
}
F.
双指针
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N=2e5+10;
#define inf 0x3f3f3f3f
void solve() {
string s,t;cin>>s>>t;
int j=0;
for(int i=0;i<s.size()&&j<t.size();i++){
if(s[i]==t[j])j++;
}
cout<<j;
}
signed main(){
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
int left=1;
//cin>>left;
while(left--){
solve();
}
}
G.
跟方格其实无关，直接判图有几个连通块
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N=1e6+10;
#define inf 0x3f3f3f3f
int m,n,k;
vector<int>g[N];
vector<int>vis(N);
void dfs(int u){
vis[u]++;
for(int i=0;i<g[u].size();i++){
int y=g[u][i];
if(vis[y])continue;
vis[y]++;
dfs(y);
}
}
void solve() {
cin>>m>>n>>k;
for(int i=0,a,b;i<k;i++){
cin>>a>>b;
g[a].push_back(b);
g[b].push_back(a);
}
int ans=0;
for(int i=1;i<=n*m;i++){
if(vis[i])continue;
ans++;
dfs(i);
}
cout<<ans;
}
signed main(){
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
int left=1;
//cin>>left;
while(left--){
solve();
}
}
H.
可以确定一定要用n/2个编号
记录下当前编号用了2次及以上的数量a，和只用了一次的数量b
说明还有n/2-a个不满足条件，设为c
若c小等于b，那么只要改动c次即可（让c对两两配对）
若c大于b，那么改动b次后，还要改动2*（c-b）次（满足两两配对）
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N=1e6+10;
#define inf 0x3f3f3f3f
void solve() {
int n;cin>>n;
map<int,int>mp;
for(int i=0,x;i<n;i++){
cin>>x;
mp[x]++;
}
int ans=0,tmp=0;
for(auto t:mp){
if(t.second>=2)ans++;
else tmp++;
}
if(ans==n/2){
cout<<"0\n";return ;
}
int t=n/2-ans;
if(t<=tmp){
cout<<t<<'\n';
return ;
}else{
cout<<(tmp+(t-tmp)*2)<<'\n';
return ;
}
}
signed main(){
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
int left=1;
//cin>>left;
while(left--){
solve();
}
}
H.
先读错题意了，wa了几发
暴力跑每一条起点到终点的路径，记录下路径上的点背经过的次数，
若次数与总路径相同，说明它是关键点
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N=1e3+10;
#define inf 0x3f3f3f3f
int n,m;
int cnt[N],path[N],vis[N];
vector<int>g[N];
int total;
int st,e;
void dfs(int u,int en,int id){
if(u==en){
for(int i=0;i<id-1;i++){
cnt[path[i]]++;
}
total++;
return ;
}
for(int i=0;i<g[u].size();i++){
int y=g[u][i];
if(!vis[y]){
vis[y]=true;
path[id]=y;
dfs(y,en,id+1);
vis[y]=false;
}
}
}
void solve() {
cin>>n>>m;
for(int i=1;i<N;i++){
vis[i]=0;cnt[i]=0;g[i].clear();total=0;
}
for(int i=0,u,v;i<m;i++){
cin>>u>>v;
g[u].push_back(v);
g[v].push_back(u);
}
cin>>st>>e;
dfs(st,e,0);
if(total==0){
cout<<-1;return ;
}
int ans=0;
for(int i=1;i<=n;i++){
if(cnt[i]==total)ans++;
}
cout<<ans;
}
signed main(){
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
int left=1;
//cin>>left;
while(left--){
solve();
}
}
I.
找规律
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N=1e3+10;
#define inf 0x3f3f3f3f
const int mod=1e9+7;
void solve() {
int n;cin>>n;
int ans=0;
for(int i=1;i<=n;i++){
ans=(ans+(i*(n-i)*(n-i))%mod)%mod;
}
cout<<ans;
}
signed main(){
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
int left=1;
//cin>>left;
while(left--){
solve();
}
}
J.
数据范围小，可以暴力跑
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N=1e3+10;
#define inf 0x3f3f3f3f
const int mod=1e9+7;
int dp[5010][5010];
void solve() {
string s;cin>>s;
int r=s.size()-1;
int ans=0;
for(int i=r;i>=0;i--){
for(int j=i+1;j<=r;j++){
if(s[i]>s[j])dp[i][j]=1;
else if(s[i]<s[j])dp[i][j]=0;
else dp[i][j]=dp[i+1][j-1];
if(dp[i][j]==1)ans++;
}
}
cout<<ans;
}
signed main(){
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
int left=1;
//cin>>left;
while(left--){
solve();
}
}
