我们的终极目标不是AC，而是获取经验
2021组队训练赛第11场
问题 A: ABB
题意
给定长度为n的字符串，在给定的字符串后面加上长度最短的字符串，使得到的字符串回文，求最少添加几个字符
考点
字符串哈希 or 马拉车算法（Manacher）
const ll inf = 1e15;
const ll INF = 0x3f3f3f3f;
const int MAXN = 410010;
char A[MAXN * 2];
int B[MAXN * 2];
void Manacher(char s[], int len)
{
int l = 0;
A[l++] = '$';//0下标存储为其他字符
A[l++] = '#';
for (int i = 0; i < len; i++)
{
A[l++] = s[i];
A[l++] = '#';
}
A[l] = 0;//空字符
int mx = 0;
int id = 0;
for (int i = 0; i < l; i++)
{
B[i] = mx > i ? min(B[2 * id - i], mx - i) : 1;
while (A[i + B[i]] == A[i - B[i]])
{
B[i]++;
}
if (i + B[i] > mx)
{
mx = i + B[i];
id = i;
}
}
return;
}
int n;
char s[MAXN];
bool judge() {
int l = 0, r = n - 1;
int flag = 0;
while (l <= r) {
if (s[l] == s[r]) {
l++;
r--;
continue;
}
else {
flag = 1;
break;
}
l++;
r--;
if (l == r) break;
}
if (flag) return false;
return true;
}
int main()
{
cin >> n;
cin >> s;
int len = (int)strlen(s);
if (judge()) {
puts("0");
return 0;
}
Manacher(s, len);
int ans = 0;
int pos = 0;
int siz = 2 * len + 1;
for (int i = 0; i < 2 * len + 1; i++) {
if (B[i] >= siz - i) {
if (i % 2) {
pos = max(pos, (B[i] - 1));
}
else
{
pos = max(pos, (B[i] - 1) / 2 * 2 + 1);
}
}
}
if (pos == 0) cout << n - 1 << endl;
else cout << n - pos << endl;
return 0;
}
/*
6
abcdcb
*/
问题 C: Bob in Wonderland
题意
将一个树形结构变成链，求出操作的次数(每次可以删除一条边)
考点
从度的角度可以很好的操作，代码很简短
const int N = 300010;
int d[N];
int main()
{
int n; scanf("%d", &n);
for (int i = 1;i < n;i++)
{
int x, y; scanf("%d%d", &x, &y);
d[x]++, d[y]++;
}
int res = 0;
for (int i = 1;i <= n;i++)
{
if(d[i] >= 2)
res += d[i] - 2;
}
cout << res;
return 0;
}
/**************************************************************
Problem: 14241
User: 我就不告诉你哈哈哈
Language: C++
Result: 正确
Time:94 ms
Memory:3196 kb
****************************************************************/
问题 F: Zeldain Garden
题意
考点
整数分块叭
计算出
x/1+x/2+x/3+x/4…+x/x即为答案，计算为下取整
typedef long long ll;
ll get(ll x)
{
ll res = 0;
for (ll l = 1, r;l <= x;l = r + 1) {
r = x / (x / l);
res += (r - l + 1) * (x / l);
}
return res;
}
int main()
{
ll l, r; scanf("%lld%lld", &l, &r);
ll ans = get(r) - get(l - 1);
printf("%lld", ans);
return 0;
}
/**************************************************************
Problem: 14244
User: 我就不告诉你哈哈哈
Language: C++
Result: 正确
Time:119 ms
Memory:2024 kb
****************************************************************/
问题 G: Light Emitting Hindenburg
题意
这个题不是我参与写的可以询问大佬
考点
据说是个贪心
代码仅从参考
虽然没代码，但是可以私信
问题 H: K == S
题意
考点
AC自动机 + 矩阵快速幂
问题 I: Ponk Warshall
/doge这个是我写的
题意
给出两个字符串，从上面字符中选取任意两个来进行交换，从上面的字符串得到下面的字符串，最少的操作步数是多少？
考点
思维 + 暴力
const ll inf = 1e15;
const ll INF = 0x3f3f3f3f;
string s, t;
map<char, int> mp;
int a[5][5];
int a1[1000007];
int a2[1000007];
int main()
{
cin >> s;
cin >> t;
mp['A'] = 1, mp['C'] = 2, mp['G'] = 3, mp['T'] = 4;
int len = s.size();
for (int i = 0; i < len; i++) {
a1[i] = mp[s[i]];
a2[i] = mp[t[i]];
}
for (int i = 0; i < len; i++) {
if (s[i] != t[i]) {
a[a1[i]][a2[i]] += 1;
}
}
ll ans = 0;
for (int i = 1; i <= 4; i++) {
for (int j = 1; j <= 4; j++) {
int mini = min(a[i][j], a[j][i]);
ans += mini;
a[i][j] -= mini;
a[j][i] -= mini;
}
}
for (int i = 1; i <= 4; i++) {
for (int j = 1; j <= 4; j++) {
for (int k = 1; k <= 4; k++) {
int mini = min(a[i][j], min(a[j][k], a[k][i]));
ans += 2 * mini;
a[i][j] -= mini;
a[j][k] -= mini;
a[k][i] -= mini;
}
}
}
//方式1：
for (int i = 1; i <= 4; i++) {
for (int j = 1; j <= 4; j++) {
for (int k = 1; k <= 4; k++) {
for (int t = 1; t <= 4; t++) {
int mini = min(a[i][j], min(min(a[j][k], a[k][t]), a[t][i]));
ans += 3 * mini;
a[i][j] -= mini;
a[j][k] -= mini;
a[k][t] -= mini;
a[t][i] -= mini;
}
}
}
}
//方式2：
for (int i = 1; i <= 4; i++) ans += 3 * a[i][2];
//ps:a[i][x],1<=x<=4均可
cout << ans << endl;
return 0;
}
/*
*/
/**************************************************************
Problem: 14247
User: 我就不告诉你哈哈哈
Language: C++
Result: 正确
Time:317 ms
Memory:15700 kb
****************************************************************/