A.Twice
题意：一个数组内选两个数相消得一分，计算最终得分
思路：用map把相等的存起来，遍历map，ans加上每个值除2
#include<bits/stdc++.h>
using namespace std;
template <typename T, typename U> T ceil(T x, U y) {return (x > 0 ? (x + y - 1) / y : x / y);}
template <typename T, typename U> T floor(T x, U y) {return (x > 0 ? x / y : (x - y + 1) / y);}
#define ll long long
#define ull unsigned long long
#define LL __int128
#define pb push_back
#define ins insert
#define fi first
#define se second
#define all(x) x.begin(),x.end()
typedef pair<ll, ll>PLL;
typedef pair<int, int>PII;
//const ll INF = 5e18;
const ll INF = 0x3f3f3f3f;
const double pi = acos(-1.0);
const int base = 131;
const ll N = 1e6 + 10;
const ll M = 6e6 + 10;
const ll MOD = 1e9 + 7;
const ll mod = 998244353;
const double eps = 0.000001;
priority_queue< ll, vector<ll>, greater<ll> > qq;
bool cmp(int a, int b) { return a > b; }
bool pri(int Su) { if (Su < 2) return false; for (int i = 2; i * i <= Su; i++) { if (Su % i == 0)return false; } return true; }
inline int gcd(int a, int b) { return b > 0 ? gcd(b, a % b) : a; }
inline int lcm(int a, int b) { return a * b / gcd(a, b); }
int lowbit(int x) { return x & (-x); }
inline ll max(ll x, ll y) { return x > y ? x : y; }
inline ll min(ll x, ll y) { return x < y ? x : y; }
ll a[N], b[N];
//double a[N], b[N];
//ull ha[N], p[N], Ha[N];
//ull hash_get(int l, int r) { return ha[r] - ha[l - 1] * p[r - l + 1]; }
ll n, m;
//cout << fixed<<setprecision()
int dx[] = { 0, 1 , 0, -1 };
int dy[] = { 1, 0 , -1, 0 };
void ManCity() {
cin >> n;
map<ll, ll>mp;
for (int i = 1; i <= n; i++) {
cin >> a[i];
mp[a[i]]++;
}
ll ans = 0;
for (auto [x, y] : mp) {
ans += y / 2;
}
cout << ans << '\n';
}
int main() {
std::ios::sync_with_stdio(false);
std::cin.tie(0);
std::cout.tie(0);
int _ = 1;
cin >> _;
while (_--) {
ManCity();
}
return 0;
}
B.Intercepted Inputs
题意:给k个数,找两个数n,m满足剩下的数个数等于n*m
思路:确定剩下的数个数一定是k-2个,找一个数能整除(k-2)就确定了一个值，再判断(k-2)/n存不存在就行
#include<bits/stdc++.h>
using namespace std;
template <typename T, typename U> T ceil(T x, U y) {return (x > 0 ? (x + y - 1) / y : x / y);}
template <typename T, typename U> T floor(T x, U y) {return (x > 0 ? x / y : (x - y + 1) / y);}
#define ll long long
#define ull unsigned long long
#define LL __int128
#define pb push_back
#define ins insert
#define fi first
#define se second
#define all(x) x.begin(),x.end()
typedef pair<ll, ll>PLL;
typedef pair<int, int>PII;
//const ll INF = 5e18;
const ll INF = 0x3f3f3f3f;
const double pi = acos(-1.0);
const int base = 131;
const ll N = 1e6 + 10;
const ll M = 6e6 + 10;
const ll MOD = 1e9 + 7;
const ll mod = 998244353;
const double eps = 0.000001;
priority_queue< ll, vector<ll>, greater<ll> > qq;
bool cmp(int a, int b) { return a > b; }
bool pri(int Su) { if (Su < 2) return false; for (int i = 2; i * i <= Su; i++) { if (Su % i == 0)return false; } return true; }
inline int gcd(int a, int b) { return b > 0 ? gcd(b, a % b) : a; }
inline int lcm(int a, int b) { return a * b / gcd(a, b); }
int lowbit(int x) { return x & (-x); }
inline ll max(ll x, ll y) { return x > y ? x : y; }
inline ll min(ll x, ll y) { return x < y ? x : y; }
ll a[N];
//double a[N], b[N];
//ull ha[N], p[N], Ha[N];
//ull hash_get(int l, int r) { return ha[r] - ha[l - 1] * p[r - l + 1]; }
ll n, m;
//cout << fixed<<setprecision()
int dx[] = { 0, 1 , 0, -1 };
int dy[] = { 1, 0 , -1, 0 };
void ManCity() {
cin >> n;
map<ll, ll>mp;
for (int i = 1; i <= n; i++) {
cin >> a[i];
mp[a[i]]++;
}
ll num = n - 2;
for (int i = 1; i <= n; i++) {
if (num % a[i] == 0) {
if (mp[num / a[i]]) {
if (num / a[i] == a[i] && mp[a[i]] == 1)continue;
cout << a[i] << ' ' << num / a[i] << '\n';
return;
}
}
}
}
int main() {
std::ios::sync_with_stdio(false);
std::cin.tie(0);
std::cout.tie(0);
int _ = 1;
cin >> _;
while (_--) {
ManCity();
}
return 0;
}
C.Superultra's Favorite Permutation
题意：找到一个n的排列满足相邻两个数之和不是质数
思路：大于2的偶数必然不是质数，只要将偶数全放一半，奇数全放一半就能满足，最小不是质数的奇数是9，所奇偶中间用4 5链接
//什么妙妙性质
#include<bits/stdc++.h>
using namespace std;
template <typename T, typename U> T ceil(T x, U y) {return (x > 0 ? (x + y - 1) / y : x / y);}
template <typename T, typename U> T floor(T x, U y) {return (x > 0 ? x / y : (x - y + 1) / y);}
#define ll long long
#define ull unsigned long long
#define LL __int128
#define pb push_back
#define ins insert
#define fi first
#define se second
#define all(x) x.begin(),x.end()
typedef pair<ll, ll>PLL;
typedef pair<int, int>PII;
//const ll INF = 5e18;
const ll INF = 0x3f3f3f3f;
const double pi = acos(-1.0);
const int base = 131;
const ll N = 1e6 + 10;
const ll M = 6e6 + 10;
const ll MOD = 1e9 + 7;
const ll mod = 998244353;
const double eps = 0.000001;
priority_queue< ll, vector<ll>, greater<ll> > qq;
bool cmp(int a, int b) { return a > b; }
bool pri(int Su) { if (Su < 2) return false; for (int i = 2; i * i <= Su; i++) { if (Su % i == 0)return false; } return true; }
inline int gcd(int a, int b) { return b > 0 ? gcd(b, a % b) : a; }
inline int lcm(int a, int b) { return a * b / gcd(a, b); }
int lowbit(int x) { return x & (-x); }
inline ll max(ll x, ll y) { return x > y ? x : y; }
inline ll min(ll x, ll y) { return x < y ? x : y; }
ll a[N];
//double a[N], b[N];
//ull ha[N], p[N], Ha[N];
//ull hash_get(int l, int r) { return ha[r] - ha[l - 1] * p[r - l + 1]; }
ll n, m;
//cout << fixed<<setprecision()
int dx[] = { 0, 1 , 0, -1 };
int dy[] = { 1, 0 , -1, 0 };
void ManCity() {
cin >> n;
vector<ll>ans;
if (n < 5) {
cout << -1 << '\n';
return;
}
for (int i = 2; i <= n; i += 2) {
if (i != 4)ans.pb(i);
}
ans.pb(4);
ans.pb(5);
for (int i = 1; i <= n; i += 2) {
if (i != 5)ans.pb(i);
}
for (auto it : ans) {
cout << it << ' ';
}
cout << '\n';
}
int main() {
std::ios::sync_with_stdio(false);
std::cin.tie(0);
std::cout.tie(0);
int _ = 1;
cin >> _;
while (_--) {
ManCity();
}
return 0;
}
D.Sharky Surfing
题意：路上有n个障碍，还有一些提升跳跃距离的装置，问想要到达L至少要捡多少个装置
思路：遍历每个障碍，如果跳跃高度小于这个障碍的长度，将在这个障碍前面的所有装置存进队列，如果这时候队列为空，则直接能判断不能越过这个障碍，否则将装置的数值加上
#include<bits/stdc++.h>
using namespace std;
template <typename T, typename U> T ceil(T x, U y) {return (x > 0 ? (x + y - 1) / y : x / y);}
template <typename T, typename U> T floor(T x, U y) {return (x > 0 ? x / y : (x - y + 1) / y);}
#define ll long long
#define ull unsigned long long
#define LL __int128
#define pb push_back
#define ins insert
#define fi first
#define se second
#define all(x) x.begin(),x.end()
typedef pair<ll, ll>PLL;
typedef pair<int, int>PII;
//const ll INF = 5e18;
const ll INF = 0x3f3f3f3f;
const double pi = acos(-1.0);
const int base = 131;
const ll N = 1e6 + 10;
const ll M = 6e6 + 10;
const ll MOD = 1e9 + 7;
const ll mod = 998244353;
const double eps = 0.000001;
priority_queue< ll, vector<ll>, greater<ll> > qq;
bool cmp(int a, int b) { return a > b; }
bool pri(int Su) { if (Su < 2) return false; for (int i = 2; i * i <= Su; i++) { if (Su % i == 0)return false; } return true; }
inline int gcd(int a, int b) { return b > 0 ? gcd(b, a % b) : a; }
inline int lcm(int a, int b) { return a * b / gcd(a, b); }
int lowbit(int x) { return x & (-x); }
inline ll max(ll x, ll y) { return x > y ? x : y; }
inline ll min(ll x, ll y) { return x < y ? x : y; }
ll l[N], r[N], x[N], v[N];
//double a[N], b[N];
//ull ha[N], p[N], Ha[N];
//ull hash_get(int l, int r) { return ha[r] - ha[l - 1] * p[r - l + 1]; }
ll n, m;
//cout << fixed<<setprecision()
int dx[] = { 0, 1 , 0, -1 };
int dy[] = { 1, 0 , -1, 0 };
void ManCity() {
ll L;
cin >> n >> m >> L;
for (int i = 1; i <= n; i++) {
cin >> l[i] >> r[i];
}
priority_queue<ll>q;
for (int i = 1; i <= m; i++) {
cin >> x[i] >> v[i];
}
ll ans = 0;
ll cnt = 1;
ll j = 1;
for (int i = 1; i <= n; i++) {
while (cnt < r[i] - l[i] + 2) {
while (j <= m && x[j] < l[i]) {
q.push(v[j]);
j++;
}
if (q.empty()) {
cout << -1 << '\n';
return;
}
cnt += q.top();
ans++;
q.pop();
}
}
cout << ans << '\n';
}
int main() {
std::ios::sync_with_stdio(false);
std::cin.tie(0);
std::cout.tie(0);
int _ = 1;
cin >> _;
while (_--) {
ManCity();
}
return 0;
}
E.Kachina's Favorite Binary String
题意：有一个字符串，每次询问l，r，回复这个区间内有多少个子序列是01的，最多询问n次，能否求得完整的字符串
思路：从1到n每次询问[1,i]，如果回复值x大于0则确定i位置上一定是1，再通过回复值x可以判断i前面是i-x-1个1后跟着若干个0，再从i+1继续询问，如果当前回复值x大于上一次询问的值，则当前i位置为1，否则为0
#include<bits/stdc++.h>
using namespace std;
template <typename T, typename U> T ceil(T x, U y) {return (x > 0 ? (x + y - 1) / y : x / y);}
template <typename T, typename U> T floor(T x, U y) {return (x > 0 ? x / y : (x - y + 1) / y);}
#define ll long long
#define ull unsigned long long
#define LL __int128
#define pb push_back
#define ins insert
#define fi first
#define se second
#define all(x) x.begin(),x.end()
typedef pair<ll, ll>PLL;
typedef pair<int, int>PII;
//const ll INF = 5e18;
const ll INF = 0x3f3f3f3f;
const double pi = acos(-1.0);
const int base = 131;
const ll N = 1e6 + 10;
const ll M = 6e6 + 10;
const ll MOD = 1e9 + 7;
const ll mod = 998244353;
const double eps = 0.000001;
priority_queue< ll, vector<ll>, greater<ll> > qq;
bool cmp(int a, int b) { return a > b; }
bool pri(int Su) { if (Su < 2) return false; for (int i = 2; i * i <= Su; i++) { if (Su % i == 0)return false; } return true; }
inline int gcd(int a, int b) { return b > 0 ? gcd(b, a % b) : a; }
inline int lcm(int a, int b) { return a * b / gcd(a, b); }
int lowbit(int x) { return x & (-x); }
inline ll max(ll x, ll y) { return x > y ? x : y; }
inline ll min(ll x, ll y) { return x < y ? x : y; }
ll a[N];
//double a[N], b[N];
//ull ha[N], p[N], Ha[N];
//ull hash_get(int l, int r) { return ha[r] - ha[l - 1] * p[r - l + 1]; }
ll n, m;
//cout << fixed<<setprecision()
int dx[] = { 0, 1 , 0, -1 };
int dy[] = { 1, 0 , -1, 0 };
void ManCity() {
cin>>n;
string s="";
ll f=0,x;
ll cnt=0;
bool flag=false,ff=false;
for(int i=2;i<=n;i++){
cout<<"? "<<1<<' '<<i<<endl;
ll x;
cin>>x;
if(x){
f=i;
cnt=x;
if(f==n){
flag=true;
f=x;
}
ff=true;
break;
}
}
if(!ff){
cout<<"! IMPOSSIBLE"<<endl;
return;
}
if(flag){
if(f==n-1){
for(int i=1;i<n;i++){
s+="0";
}
s+="1";
cout<<"! "<<s<<endl;
return;
}
for(int i=1;i<n-f;i++){
s+="1";
}
for(int i=n-f;i<n;i++){
s+="0";
}
s+="1";
cout<<"! "<<s<<endl;
return;
}
for(int i=1;i<f-cnt;i++){
s+="1";
}
for(int i=f-cnt;i<f;i++){
s+="0";
}
s+="1";
cnt=x;
for(int i=f+1;i<=n;i++){
cout<<"? "<<1<<' '<<i<<endl;
cin>>x;
if(x>cnt)s+="1";
else s+="0";
cnt=x;
}
cout<<"! "<<s<<endl;
}
int main() {
std::ios::sync_with_stdio(false);
std::cin.tie(0);
std::cout.tie(0);
int _ = 1;
cin >> _;
while (_--) {
ManCity();
}
return 0;
}
