（GPLT）上海理工大学校内选拔赛（重现赛）
A.
记录最大值和个数，和次大值
遍历数组，不是最大值，输出该数与最大值的和
是最大值，则讨论最大值个数，若只有一个，输出该数与次大值的和，否则输出最大值的两倍
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N=1e5+10;
void solve() {
int n;cin>>n;
vector<int>a(n+1);
int ma=-1,cnt_ma=0,tm=-1;
for(int i=1;i<=n;i++){
cin>>a[i];
if(a[i]>ma){
tm=ma;
ma=a[i];
cnt_ma=1;
}else if(a[i]==ma){
cnt_ma++;
}else if(a[i]>tm){
// cout<<"tm="<<a[i]<<'\n';
tm=a[i];
}
}
// cout<<ma<<' '<<tm;
for(int i=1;i<=n;i++){
if(a[i]<ma){
cout<<a[i]+ma;
}else if(a[i]==ma){
if(cnt_ma>1){
cout<<2*ma;
}else cout<<a[i]+tm;
}
if(i!=n)cout<<' ';
}
}
signed main(){
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
int left=1;
//cin>>left;
while(left--){
solve();
}
}
B.
快速幂即可
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N=1e5+10;
int ksm(int a,int b,int p){
int ans=1;
while(b){
if(b&1)ans=(ans*a)%p;
a=(a*a)%p;
b/=2;
}
return ans%p;
}
void solve() {
int n;cin>>n;
cout<<ksm(n,n,n+2);
}
signed main(){
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
int left=1;
//cin>>left;
while(left--){
solve();
}
}
C.
从字符串里处理出字符串并计数，用map实现
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N=1e5+10;
void solve() {
string s;getline(cin,s);
//cout<<s;
map<string,int>mp;
string tmp="";
int sum=0;
for(int i=0;i<s.size();i++){
if(s[i]>='a'&&s[i]<='z'||s[i]>='A'&&s[i]<='Z')tmp+=s[i];
else if(s[i]>='0'&&s[i]<='9')sum=sum*10+(s[i]-'0');
else if(s[i]==' '){
if(sum){
mp[tmp]+=sum;
sum=0;
tmp="";
}
}
if(i==s.size()-1){
mp[tmp]+=sum;
}
}
int tt;cin>>tt;
string t;
while(tt--){
cin>>t;
cout<<mp[t]<<'\n';
}
}
signed main(){
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
int left=1;
//cin>>left;
while(left--){
solve();
}
}
D.
K数列就是进制的定义
类比二进制串，想象一个k进制串
把n先转成二进制串，再转为k进制串即可
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N=1e5+10;
const int mod=1e9+7;
void solve() {
int n,k;cin>>n>>k;
int ans=0,t=1;
while(n){
if(n&1)ans=(ans+t)%mod;
t=(t*k)%mod;
n/=2;
}
cout<<ans;
}
signed main(){
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
int left=1;
//cin>>left;
while(left--){
solve();
}
}
E.
bfs
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N=300+10;
int g[N][N];
int n,m,k,a,b,c,d;
int vis1[N][N],vis2[N][N];
struct node{
int x,y,dep;
};
int dx[10]={-2,-2,-1,-1,2,2,1,1};
int dy[10]={1,-1,2,-2,1,-1,2,-2};
int cx[10]={-1,-1,0,0,1,1,0,0};
int cy[10]={0,0,1,-1,0,0,1,-1};
int bfs1(){
queue<node>q;
while(q.size())q.pop();
if(a==c&&b==d)return 0;
q.push({a,b,0});
vis1[a][b]=1;
while(q.size()){
auto t=q.front();q.pop();
for(int i=0;i<8;i++){
int xx=t.x+dx[i],yy=t.y+dy[i];
if(xx>=1&&xx<=n&&yy>=1&&yy<=m){
if(vis1[xx][yy]||g[xx][yy])continue;
vis1[xx][yy]=1;
if(xx==c&&yy==d)return t.dep+1;
q.push({xx,yy,t.dep+1});
}
}
}
return -1;
}
int bfs2(){
queue<node>q;
while(q.size())q.pop();
if(a==c&&b==d)return 0;
q.push({a,b,0});
vis2[a][b]=1;
while(q.size()){
auto t=q.front();q.pop();
for(int i=0;i<8;i++){
int xx=t.x+dx[i],yy=t.y+dy[i];
int tx=t.x+cx[i],ty=t.y+cy[i];
if(xx<1||xx>n||yy<1||yy>m)continue;
if(g[xx][yy]||g[tx][ty])continue;
if(vis2[xx][yy])continue;
vis2[xx][yy]=1;
if(xx==c&&yy==d)return t.dep+1;
q.push({xx,yy,t.dep+1});
}
}
return -1;
}
void solve() {
for(int i=1;i<N;i++){
for(int j=1;j<N;j++){
g[i][j]=0;vis1[i][j]=0;vis2[i][j]=0;
}
}
cin>>n>>m>>k>>a>>b>>c>>d;
while(k--){
int x,y;cin>>x>>y;
g[x][y]=1;
}
cout<<bfs1()<<' '<<bfs2()<<'\n';
}
signed main(){
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
int left=1;
cin>>left;
while(left--){
solve();
}
}
F.
最大配最小
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N=1e5+10;
void solve() {
int n;cin>>n;
vector<int>a(n+1),b(n+1);
for(int i=1;i<=n;i++)cin>>a[i];
for(int i=1;i<=n;i++)cin>>b[i];
sort(a.begin()+1,a.end());
sort(b.begin()+1,b.end());
int ans=0;
for(int i=1;i<=n;i++){
ans+=a[i]*b[n-i+1];
}
cout<<ans<<'\n';
}
signed main(){
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
int left=1;
cin>>left;
while(left--){
solve();
}
}
J.
dij板子
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N=1e5+10;
vector<pair<int,int>>g[N];
#define inf 0x3f3f3f3f3f3f3f3f
int dist[N];
bool st[N];
priority_queue<pair<int,int>,vector<pair<int,int>>,greater<pair<int,int>>> q;
void dij(){
memset(dist,inf,sizeof(dist));
dist[1]=0;
q.push({0,1});
while(q.size()){
auto k=q.top();q.pop();
int ver=k.second,distance=k.first;
if(st[ver])continue;
st[ver]=true;
for(int i=0;i<g[ver].size();i++){
int j=g[ver][i].first;
if(dist[j]>distance+g[ver][i].second){
dist[j]=distance+g[ver][i].second;
q.push({dist[j],j});
}
}
}
}
void solve() {
int n,m;cin>>n>>m;
for(int i=1,v,w,d;i<=m;i++){
cin>>v>>w>>d;
g[v].push_back({w,d});
g[w].push_back({v,d});
}
dij();
int tt;cin>>tt;
while(tt--){
int s,t;cin>>s>>t;
cout<<dist[s]+dist[t]<<'\n';
}
}
signed main(){
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
int left=1;
//cin>>left;
while(left--){
solve();
}
}
