T1：Yay!
模拟
代码实现
#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)
using namespace std;
int main() {
string s;
cin >> s;
vector<int> cnt(256);
for (char c : s) cnt[c]++;
char one = '?';
for (char c = 'a'; c <= 'z'; ++c) {
if (cnt[c] == 1) one = c;
}
rep(i, s.size()) {
if (s[i] == one) {
cout << i+1 << '\n';
}
}
return 0;
}
T2：Which is ahead?
模拟
代码实现
#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)
using namespace std;
int main() {
int n;
cin >> n;
vector<int> p(n);
rep(i, n) cin >> p[i];
vector<int> x(n+1);
rep(i, n) x[p[i]] = i;
int q;
cin >> q;
rep(qi, q) {
int a, b;
cin >> a >> b;
if (x[a] < x[b]) cout << a << '\n';
else cout << b << '\n';
}
return 0;
}
T3：Many Replacement
只需记录每种字符最后变成了哪个字符即可
代码实现
#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)
using namespace std;
int main() {
int n, q;
string s;
cin >> n >> s >> q;
string t;
rep(i, 26) t += 'a'+i;
rep(qi, q) {
char c, d;
cin >> c >> d;
rep(i, 26) {
if (t[i] == c) t[i] = d;
}
}
rep(i, n) {
int j = s[i]-'a';
s[i] = t[j];
}
cout << s << '\n';
return 0;
}
T4：Square Pair
可以先将每个数中的最大的平方因子除掉，然后就变成简单的组合计数问题了
代码实现
#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
int main() {
int n;
cin >> n;
vector<int> a(n);
rep(i, n) cin >> a[i];
rep(i, n) {
int x = a[i];
for (int j = 2; ; ++j) {
int y = j*j;
if (y > x) break;
while (x%y == 0) x /= y;
}
a[i] = x;
}
ll ans = 0;
int zero = 0;
rep(i, n) {
if (a[i] == 0) {
zero++;
}
}
ans += (ll)zero*(zero-1)/2;
map<int, int> mp;
for (int x : a) {
if (x == 0) continue;
ans += zero;
ans += mp[x];
mp[x]++;
}
cout << ans << '\n';
return 0;
}