比赛地址：https://codeforces.com/contest/1682
A
思路：从中间往左右两侧入手，判断有组相同。
奇数中间的就是目标字符goal，往左右两边检查是否和goal相同。
偶数中间两个就是目标字符goal，往左右两边检查是否和goal相同
#define LOCAL
/*字符串 长n
计算i的数量，删掉了Si后仍然是回文序列
每一次只移除掉一个，看能移除掉几个满足回文
*/
#include <bits/stdc++.h>
#define rep(i, a, b) for(int i = a; i <= b; i ++)
#define endl '\n'
using namespace std;
const int N = 1e9 + 5;
void s(){
int n ;cin >> n;
string a ; cin >> a;
a = " " + a;
long long ans = 0;
if(n % 2 == 1){
//odd
ans = 1;
char goal = a[n/2+1];
for(int i = n / 2 + 1 - 1, j = n / 2 + 1 + 1; i >= 1 && j <= n ; i --, j ++){
int tag = 1;
if(a[i] == goal){
ans++;
}else tag = 0;
if(a[j] == goal){
ans++;
}else tag = 0;
if(!tag) break;
}
}else{
//even
if(a[n/2] != a[n/2 + 1]) {
cout << ans << endl;
return ;
}
char goal = a[n/2];
for(int i = n / 2, j = n / 2 + 1; i >= 1 && j <= n; i -- , j ++){
int tag = 1;
if(a[i] == goal){
ans++;
}else tag = 0;
if(a[j] == goal){
ans++;
}else tag = 0;
if(!tag) break;
}
}
cout << ans << endl;
}
int main(){
cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);
#ifdef LOCAL
cout << "\n记得更改input.txt!!!\n";
FILE *p = fopen("input.txt", "a+");
fclose(p);
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int t; cin >>t;while(t-- )s();
return 0;
}
B
借鉴了https://zhuanlan.zhihu.com/p/518427683
/*
给你一个序列p 0---n-1只出现一次。最初序列没有排序
按位&
p满足xsort
*/
#include <bits/stdc++.h>
#define rep(i, a, b) for(int i = a; i <= b; i ++)
#define endl '\n'
using namespace std;
const int N = 2e5 + 5;
int a[N], q[N];
void s(){
int n; cin >> n;
int x = -1;
rep(i, 0 , n-1){
cin >> a[i];
if(a[i] != i) x &= a[i];
}
cout << x << endl;
}
int main(){
cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);
#ifdef LOCAL
cout << "\n记得更改input.txt!!!\n";
FILE *p = fopen("input.txt", "a+");
fclose(p);
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int t; cin >> t;
while(t--) s();
return 0;
}
C
借鉴同上B
/*n个正数
lis(a) = 最长严格增加子序列
定义：a撇 为a倒叙后的序列
美丽数组被定为min（lis （a） ， lis（a‘））
你的任务就是去决定美丽数组的最大值。你能够重新任意排序
*/
#include <bits/stdc++.h>
#define rep(i, a, b) for(int i = a; i <= b; i ++)
#define endl '\n'
using namespace std;
const int N = 2e5 + 5;
int q[N], tt;
void s(){
int n ; cin >> n;
int a[N] = {};
map<int, int> mp;
rep(i, 1, n) cin >> a[i], mp[a[i]] ++;
sort(a + 1, a + 1 + n);
int len = unique(a+1, a+1+n) - a - 1;
int one = 0, two = 0;
rep(i,1,len){
if(mp[a[i]] >= 2) two++;
if(mp[a[i]] == 1) one++;
}
cout << two + (one + 1) / 2 << endl;
}
int main(){
cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);
#ifdef LOCAL
cout << "\n记得更改input.txt!!!\n";
FILE *p = fopen("input.txt", "a+");
fclose(p);
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int t; cin >>t; while(t--) s();
return 0;
}
D
借鉴同上B
/*
n个结点在一个圆圈里，从1---n顺时针顺序
给你一个二进制string长度n
你的任务就是构造一棵树，满足下面两个或者树不存在：
每一个结点i，结点的度如果si=0是偶数如果si = 1则为奇数
在圆圈里没有两条边连接。边允许连接圆圈
所有的边只是画一条线（无向边）
一棵树有n-1条边
*/
#include <bits/stdc++.h>
#define rep(i, a, b) for(int i = a; i <= b; i ++)
#define endl '\n'
using namespace std;
const int N = 1e9 + 5;
void s() {
int n;
string s;
cin >> n >> s;
int onecnt = 0;
rep(i,0,n-1) onecnt += s[i] - '0';
if(onecnt&1 || onecnt == 0){
cout << "NO\n";return ;
}
cout << "YES\n";
int pos = s.find('1');
int last = s.find_last_of('1');
int pre = pos;//pre前
for(int i = pos + 1; i != last ; i ++){
if(s[i] == '1'){
cout << pre + 1 << " " << i + 1 << '\n';
pre = pos;
}
else {
cout << pre + 1 << " " << i + 1 << '\n';
pre = i;
}
}
for(int i = (pos - 1 + n) % n ; i != last -1; i = (i + n - 1) % n){
cout << pre + 1 << " " << i + 1 << '\n';
pre = i;
}
}
int main(){
cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);
#ifdef LOCAL
cout << "\n记得更改input.txt!!!\n";
FILE *p = fopen("input.txt", "a+");
fclose(p);
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int t; cin >> t; while(t--) s();
return 0;
}
