Educational Codeforces Round 75 (Rated for Div. 2)
A. Broken Keyboard
-
思路：最开始没搞清楚题意瞎搞 只需要判断前后两连续字符是否相同 不同则将前一个存入vector中 排序去个重输出即可
-
AC代码
#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
ll mult_mod(ll x, ll y, ll mod){
return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}
ll pow_mod(ll a, ll b, ll p){
ll res = 1;
while (b){
if (b & 1)
res = mult_mod(res, a, p);
a = mult_mod(a, a, p);
b >>= 1;
}
return res % p;
}
ll gcd(ll a, ll b){
return b ? gcd(b, a % b) : a;
}
int t;
string s;
vector<char> res;
int main(){
#ifndef ONLINE_JUDGE
freopen("my_in.txt", "r", stdin);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> t;
while (t -- ){
cin >> s;
if (s.length() == 1){
cout << s[0] << "
";
continue;
}
res.clear();
int i = 1;
if (s[0] != s[1])
res.push_back(s[0]);
else
i = 2;
for (; i < s.length(); i ++ ){
if (s[i] == s[i + 1])
i ++ ;
else
res.push_back(s[i]);
}
sort(res.begin(), res.end());
res.erase(unique(res.begin(), res.end()), res.end());
for (int i = 0; i < res.size(); i ++ )
cout << res[i];
cout << "
";
}
return 0;
}
B. Binary Palindromes
-
思路：当时想了一个错误的贪心思路 直接统计0 1的个数 然后先分配偶数长度的串 然后分配奇数长度的串 写了很长的模拟 最后wa了 只要有奇数长度的串或者0或1的个数是奇数就输出(n) 否则输出(n - 1)就行辽
-
AC代码
#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
ll mult_mod(ll x, ll y, ll mod){
return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}
ll pow_mod(ll a, ll b, ll p){
ll res = 1;
while (b){
if (b & 1)
res = mult_mod(res, a, p);
a = mult_mod(a, a, p);
b >>= 1;
}
return res % p;
}
ll gcd(ll a, ll b){
return b ? gcd(b, a % b) : a;
}
const int N = 60;
int q, n, cnt0, cnt1, cnt, len;
string s;
int main(){
#ifndef ONLINE_JUDGE
freopen("my_in.txt", "r", stdin);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> q;
while (q -- ){
cnt = 0, cnt0 = 0, cnt1 = 0;
cin >> n;
for (int i = 1; i <= n; i ++ ){
cin >> s;
len = s.size();
if (len & 1)
cnt ++ ;
for (int j = 0; j < len; j ++ ){
if (s[j] == '0')
cnt0 ++ ;
else
cnt1 ++ ;
}
}
// debug cout << cnt0 << " " << cnt1 << "
";
if (cnt || (cnt0 % 2 == 0))
cout << n << "
";
else
cout << n - 1 << "
";
}
return 0;
}
C. Minimize The Integer
-
思路：奇数串儿和偶数串儿的顺序是不会改变的 用俩string分别存奇数串儿和偶数串儿 然后分别从前往后比较大小 小的存在前面即可
-
AC代码
#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
ll mult_mod(ll x, ll y, ll mod){
return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}
ll pow_mod(ll a, ll b, ll p){
ll res = 1;
while (b){
if (b & 1)
res = mult_mod(res, a, p);
a = mult_mod(a, a, p);
b >>= 1;
}
return res % p;
}
ll gcd(ll a, ll b){
return b ? gcd(b, a % b) : a;
}
int t;
string s;
int main(){
#ifndef ONLINE_JUDGE
freopen("my_in.txt", "r", stdin);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> t;
while (t -- ){
cin >> s;
string ans;
string odd, even;
for (int i = 0; i < s.length(); i ++ ){
if (s[i] & 1)
odd.push_back(s[i]);
else
even.push_back(s[i]);
}
for (int k = 0, i = 0, j = 0; k < s.length(); k ++ ){
if (i == odd.size()){
ans.push_back(even[j]);
j ++ ;
}
else if (j == even.size()){
ans.push_back(odd[i]);
i ++ ;
}
else{
if (odd[i] >= even[j]){
ans.push_back(even[j]);
j ++ ;
}
else{
ans.push_back(odd[i]);
i ++ ;
}
}
}
cout << ans << "
";
}
return 0;
}
D. Salary Changing
-
思路：贪心 + 二分 先以左端点从大到小排序 然后二分找中间值即可(
cmp函数l写成r debug已知没找到 然后疯狂wa -
AC代码
#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
ll mult_mod(ll x, ll y, ll mod){
return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}
ll pow_mod(ll a, ll b, ll p){
ll res = 1;
while (b){
if (b & 1)
res = mult_mod(res, a, p);
a = mult_mod(a, a, p);
b >>= 1;
}
return res % p;
}
ll gcd(ll a, ll b){
return b ? gcd(b, a % b) : a;
}
const int N = 2e5 + 10;
const int INF = 0x3f3f3f3f;
int t, n;
ll s, ans;
struct Seg{
ll l, r;
}seg[N];
inline bool cmp(Seg s1, Seg s2){
if (s1.l == s2.l)
return s1.r > s2.r;
return s1.l > s2.l;
}
inline bool check(ll x){
ll cnt = n / 2 + 1, sum = 0;
for (int i = 1; i <= n; i ++ ){
if (seg[i].l <= x && seg[i].r >= x && cnt){
sum += x;
cnt -- ;
}
else{
sum += seg[i].l;
if (seg[i].l > x)
cnt -- ;
}
}
if (cnt)
return false;
return sum <= s;
}
inline ll calc(){
ll l = seg[n / 2 + 1].l + 1, r = s;
ll mid;
while (l <= r){
mid = (l + r) >> 1;
if (check(mid))
l = mid + 1;
else
r = mid - 1;
}
return r;
}
int main(){
#ifndef ONLINE_JUDGE
freopen("my_in.txt", "r", stdin);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> t;
while (t -- ){
cin >> n >> s;
for (int i = 1; i <= n; i ++ )
cin >> seg[i].l >> seg[i].r;
sort(seg + 1, seg + n + 1, cmp);
ans = calc();
cout << ans << "
";
}
return 0;
}