Codeforces Round #603 (Div. 2)
Codeforces Round #603 (Div. 2)
A. Sweet Problem
-
思路：水题 当时有点降智 贡献一发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;
}
int t;
ll a, b, c, max_, min_, sum, ans;
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 >> a >> b >> c;
max_ = max(a, max(b, c));
min_ = min(a, min(b, c));
sum = a + b + c;
ans = sum - max_ - min_;
if (min_ + ans > max_)
cout << max_ + (min_ + ans - max_) / 2 << "\n";
else
cout << min_ + ans << "\n";
}
return 0;
}
B. PIN Codes
-
思路：由于每次最多只有十个数 故只用管最后一位从0到9整就行了
-
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 = 20;
int t, n, tot;
string p[N], ans[N];
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 -- ){
tot = 0;
cin >> n;
for (int i = 1; i <= n; i ++ )
cin >> p[i];
for (int i = 1; i <= n; i ++ ){
for (int j = 1; j <= n; j ++ ){
if (i == j)
continue;
if (p[i] == p[j]){
while (true){
bool flag = false;
if (p[i][3] != '9')
p[i][3] = char((int)p[i][3] + 1);
else
p[i][3] = '0';
for (int k = 1; k <= n; k ++ ){
if (i == k)
continue;
if (p[i] == p[k]){
flag = true;
break;
}
}
if (!flag)
break;
}
tot ++ ;
break;
}
}
}
cout << tot << "\n";
for (int i = 1; i <= n; i ++ )
cout << p[i] << "\n";
}
return 0;
}
C. Everyone is a Winner!
-
思路：分块
-
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 = 1e7 + 10;
int t, n, tot, res, cnt, tmp;
int ans[N];
bool flag;
int primes[N];
bool vis[N];
void Init(){
for (int i = 2; i < N; i ++ ){
if (!vis[i])
primes[tot ++ ] = i;
for (int j = 0; j < tot && i * primes[j] <= n; j ++ ){
vis[i * primes[j]] = true;
if (!(i % primes[j]))
break;
}
}
}
int main(){
#ifndef ONLINE_JUDGE
freopen("my_in.txt", "r", stdin);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
Init();
cin >> t;
while (t -- ){
tmp = -1, cnt = 1, res = 0;
flag = false;
cin >> n;
ans[0] = n;
for (int i = 0; i <= tot - 1; i ++ ){
if (primes[i - 1] > n)
break;
if (tmp == n / primes[i]){
flag = true;
break;
}
else{
ans[cnt ++ ] = n / primes[i];
tmp = n / primes[i];
}
}
if (flag)
res += tmp;
res += cnt;
cout << res << "\n";
if (flag)
for (int i = 0; i <= tmp - 1; i ++ )
cout << i << " ";
for (int i = cnt - 1; i > 0; i -- )
cout << ans[i] << " ";
cout << ans[0] << "\n";
}
return 0;
}
D. Secret Passwords
-
思路：并查集维护即可
-
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;
}
const int N = 50;
int n, res, ans, tmp;
int fa[N], flag1[N], flag2[N];
string s;
int find(int x){
if (fa[x] == x)
return x;
return fa[x] = find(fa[x]);
}
void Init(){
for (int i = 1; i < N; i ++ )
fa[i] = i;
}
int main(){
#ifndef ONLINE_JUDGE
freopen("my_in.txt", "r", stdin);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
Init();
cin >> n;
for (int i = 1; i <= n; i ++ ){
cin >> s;
tmp = s[0] - 'a' + 1;
flag1[tmp] = true;
res = find(tmp);
for (int j = 1; j < s.length(); j ++ ){
tmp = s[j] - 'a' + 1;
fa[tmp] = res;
flag1[tmp] = true;
}
}
for (int i = 1; i <= 26; i ++ ){
fa[i] = find(fa[i]);
if (!flag2[fa[i]] && flag1[i]){
flag2[fa[i]] = true;
ans ++ ;
}
}
cout << ans << "\n";
return 0;
}
