A.买爱心气球
原题链接 : 登录—专业IT笔试面试备考平台_牛客网
博弈论 :
#include <iostream>
using namespace std;
int t,n,m;
string s1 = "Alice",s2 = "Bob";
int main() {
cin>>t;
while(t--){
cin>>n>>m;
if (n % 3 == 0) {
cout << s1 << endl;
continue;
}
if (n % 3 == 1) {
cout << (m != 1 ? "Alice\n" : "Bob\n");
continue;
}
if (n % 3 == 2) {
cout << (m != 2 ? "Alice\n" : "Bob\n");
continue;
}
}
return 0;
}B.亚托莉 -我挚爱的时光-
原题链接 : 登录—专业IT笔试面试备考平台_牛客网
题面 : 
思路 : 模拟
#include "bits/stdc++.h"
using namespace std;
using i64 = long long;
int main() {
int n;
cin >> n;
auto useless = cin.get();
vector<string> a(n);
for (int i = 0 ;i < n ;i ++) {
getline(cin, a[i]);
}
string c[] = {
"sudo pacman -S ",
"pacman -R ",
"pacman -Rscn ",
"sudo rm -rf /*"
};
unordered_set<string> st[2];
for (int i = 0; i < n; i++) {
auto check = [&](int j) {
int x = a[i].size(), y = c[j].size();
if (x >= y && a[i].substr(0, y) == c[j]) {
return true;
}
return false;
};
if (check(0)) {
auto s = a[i].substr(c[0].size());
st[0].insert(s);
st[1].insert(s);
} else if (check(1)) {
auto s = a[i].substr(c[1].size());
st[0].erase(s);
} else if (check(2)) {
auto s = a[i].substr(c[2].size());
st[0].erase(s);
st[1].erase(s);
} else if (check(3)) {
cout << "wuwuwu\n";
return 0;
} else {
auto o = a[i][0] - '0' - 1;
auto s = a[i].substr(2);
cout << (st[o].count(s) ? "yes" : "no") << '\n';
}
}
return 0;
}D.01分数规划
链接 : 登录—专业IT笔试面试备考平台_牛客网
题面 : 
思路 : 分两种情况，?全替换为0,或全替换为1.
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
string s,t; int T,n;
void solve(){
cin>>n; cin>>s;
s=' '+s; t=s;
for(int i=1;i<=n;i++) if(s[i]=='?') s[i]='1';
for(int i=1;i<=n;i++) if(t[i]=='?') t[i]='0';
int ans = 0;
for(int i=1;i<=n;i++){
int j =i;
while(j<=n&&s[j]==s[i]) j++; j--;
ans = max(ans,j-i+1);
i=j;
}
for(int i=1;i<=n;i++){
int j = i;
while(j<=n&&t[j]==t[i]) j++; j--;
ans = max(ans,j-i+1);
i=j;
}
cout<<ans<<endl;
return ;
}
int main(){
cin>>T;
while(T--) solve();
return 0;
}I.双指针
原题链接 : 登录—专业IT笔试面试备考平台_牛客网
题面 : 
思路 : 用hash表来存每次的a/b，然后ans加上之前在map中出现相反数的次数
#include<bits/stdc++.h>
#define IOS ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define endl '\n'
using namespace std;
typedef long long LL;
const int N = 2e5+10;
unordered_map<double,int> mp;
int a[N],b[N],n;
inline void solve(){
mp.clear();
cin>>n;
LL ans = 0;
for(int i=1;i<=n;i++) cin>>a[i];
for(int j=1;j<=n;j++) cin>>b[j];
for(int i=1;i<=n;i++){
double x = 1.0*a[i]/b[i],y=1.0*b[i]/a[i];
ans += mp[x];
mp[y]++;
}
cout<<ans<<endl;
}
int main()
{
IOS
int _;
cin >> _;
while(_ --) solve();
return 0;
}
// ai*aj =bi*bj
// ai / bi = bj / aj
// 1 2 1 1/2 3/2J.树上dp
原题链接 : 登录—专业IT笔试面试备考平台_牛客网
题面 : 
思路 : 贪心，权值大的考虑放在深度大的地方
#include<bits/stdc++.h>
#define IOS ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define endl '\n'
using namespace std;
typedef pair<int, int> PII;
typedef long long ll;
const int N = 200010, M = N * 2;
int h[N], e[M], ne[M], idx;
int val[N], d[N];
bool st[N];
inline void add(int a, int b)
{
e[idx] = b, ne[idx] = h[a], h[a] = idx ++;
}
void dfs(int u)
{
st[u] = true;
for(int i = h[u]; i != -1; i = ne[i])
{
int j = e[i];
if(st[j])continue;
d[j] = d[u] + 1;
dfs(j);
}
}
inline void solve()
{
memset(h, -1, sizeof h);
idx = 0;
int n;
cin >> n;
for(int i = 1; i < n; i ++)
{
int a, b;
cin >> a >> b;
add(a, b), add(b, a);
}
for(int i = 1; i <= n; i ++)
{
cin >> val[i];
st[i] = false;
}
d[1] = 1;
dfs(1);
sort(d + 1, d + 1 + n);
sort(val + 1, val + 1 + n);
ll ans = 0;
for(int i = 1; i <= n; i ++)
{
ans += (ll)val[i] * (ll)d[i];
}
cout << ans << endl;
}
int main()
{
IOS
int _;
cin >> _;
while(_ --)
{
solve();
}
return 0;
}
