AtCoder Beginner Contest 272
A - Integer Sum
签到题，求和就好
#include<bits/stdc++.h>
using namespace std;
int read() {
int x = 0, f = 1, ch = getchar();
while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
if (ch == '-') f = -1, ch = getchar();
while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
return x * f;
}
int32_t main() {
int n = read() , sum = 0;
for( int x ; n ; n -- )
x = read() , sum += x;
cout << sum << '\n';
return 0;
}
B - Everyone is Friends
对于每一组，\(O(n^2)\)的枚举两个人，然后标记他们的关系。最后遍历任意两人之间的关系
#include<bits/stdc++.h>
using namespace std;
int read() {
int x = 0, f = 1, ch = getchar();
while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
if (ch == '-') f = -1, ch = getchar();
while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
return x * f;
}
const int N = 105;
int a[N];
bool v[N][N];
int32_t main() {
int n = read() , m = read();
for( int k ; m ; m -- ){
k = read();
for( int j = 1 ; j <= k ; j ++ )
a[j] = read();
for( int i = 1 ; i <= k ; i ++ )
for( int j = 1 ; j < i ; j ++ )
v[ a[i] ][ a[j] ] = v[ a[j] ][ a[i] ] = 1;
}
for( int i = 1 ; i <= n ; i ++ )
for( int j = 1 ; j < i ; j ++ ){
if( v[i][j] ) continue;
cout << "No\n";
return 0;
}
cout << "Yes\n";
return 0;
}
C - Max Even
找到奇数的最大和次大，偶数的最大和次大。然后取max
#include<bits/stdc++.h>
#define int long long
using namespace std;
int read() {
int x = 0, f = 1, ch = getchar();
while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
if (ch == '-') f = -1, ch = getchar();
while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
return x * f;
}
const int N = 105;
int a = INT_MIN , b = INT_MIN , c = INT_MIN , d = INT_MIN;
int32_t main() {
int n = read();
for( int x ; n ; n -- ){
x = read();
if( x & 1 ){
if( x > a ) b = a , a = x;
else if( x > b ) b = x;
}
else{
if( x > c ) d = c , c = x;
else if( x > d ) d = x;
}
}
int res = max( a+b , c + d );
if( res < 0 ) cout << -1;
else cout << res << "\n";
return 0;
}
D - Root M Leaper
算出所有移动距离为\(\sqrt m\)的移动向量，然后就是bfs一下啊就好
#include<bits/stdc++.h>
//#define int long long
using namespace std;
int read() {
int x = 0, f = 1, ch = getchar();
while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
if (ch == '-') f = -1, ch = getchar();
while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
return x * f;
}
const int N = 405;
int n , m , dis[N][N];
vector< pair<int,int> > e;
int32_t main() {
n = read() , m = read();
for( int i = 0 , j ; i <= m ; i ++ ){
j = (int)sqrt( m - i * i );
if( i * i + j * j == m ) {
e.emplace_back( i , j );
e.emplace_back( i , -j );
e.emplace_back( -i , j );
e.emplace_back( -i , -j );
}
}
fill( dis[0] , dis[0] + N * N , INT_MAX );
dis[1][1] = 0;
queue< pair<int,int> > q;
q.push( { 1 , 1 } );
while( !q.empty() ){
auto [ x , y ] = q.front() ; q.pop();
for( auto [ dx , dy ] : e ){
int fx = x + dx , fy = y + dy;
if( fx < 1 || fx > n || fy < 1 || fy > n || dis[fx][fy] <= dis[x][y] + 1 ) continue;
dis[fx][fy] = dis[x][y] + 1 , q.push( { fx , fy });
}
}
for( int i = 1 ; i <= n ; i ++ )
for( int j = 1 ; j <= n ; j ++ )
cout << ( dis[i][j] == INT_MAX ? -1 : dis[i][j] ) << ( j == n ? '\n' : ' ' );
return 0;
}
E - Add and Mex
每一个数只能加到大于等于0时才有贡献，加到大于n时就不会再有贡献，所以二分出最小的可以贡献的值，然后暴力加到没有贡献，再此过从中保存下所有可能产生贡献的值。
对于每次操作，暴力一下循环一下就可以找到mex。
#include<bits/stdc++.h>
#define int long long
using namespace std;
int read() {
int x = 0, f = 1, ch = getchar();
while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
if (ch == '-') f = -1, ch = getchar();
while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
return x * f;
}
const int N = 2e5+5;
int n , m;
vector<int> p[N];
int search( int i , int x ){
int l = 1 , r = m , res , mid ;
while( l <= r ){
mid = ( l + r ) >> 1;
if( mid * i + x >= 0 ) res = mid , r = mid - 1;
else l = mid + 1;
}
return res;
}
int32_t main() {
n = read() , m = read();
for( int i = 1 , x , t ; i <= n ; i ++ ){
x = read() ;
if( m * i + x < 0 ) continue;
t = search( i , x ) ;
for( int j = t ; j <= m ; j ++ ){
if( j * i + x > n ) break;
p[j].push_back( j * i + x ) ;
}
}
for( int i = 1 , res ; i <= m ; i ++ ){
res = 0;
sort( p[i].begin() , p[i].end() );
for( auto it : p[i] ){
if( it == res ) res ++;
else if( it < res ) continue;
else break;
}
cout << res << "\n";
}
return 0;
}
AtCoder Beginner Contest 272
