CodeForces1230

CodeForces1230A

CodeForces1230A
题目并不难,只需要注意不要犯智障错误即可.
智障错误包括但不限于:以为要两两一组分两组和判断两部分是否相等时总和与总和-当前集合比较...

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <string>
#include <vector>
#include <queue>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
#define MEM(x,y) memset ( x , y , sizeof ( x ) )
#define rep(i,a,b) for (int i = (a) ; i <= (b) ; ++ i)
#define per(i,a,b) for (int i = (a) ; i >= (b) ; -- i)
#define pii pair < int , int >
#define one first
#define two second
#define rint read<int>
#define int long long
#define pb push_back
#define db double

using std::queue ;
using std::set ;
using std::pair ;
using std::max ;
using std::min ;
using std::priority_queue ;
using std::vector ;
using std::swap ;
using std::sort ;
using std::unique ;
using std::greater ;

template < class T >
    inline T read () {
        T x = 0 , f = 1 ; char ch = getchar () ;
        while ( ch < '0' || ch > '9' ) {
            if ( ch == '-' ) f = - 1 ;
            ch = getchar () ;
        }
       while ( ch >= '0' && ch <= '9' ) {
            x = ( x << 3 ) + ( x << 1 ) + ( ch - 48 ) ;
            ch = getchar () ;
       }
   return f * x ;
}

int can[6] , sum ;

inline bool check (int x , int y) { return ( sum - can[x] - can[y] ) == ( can[x] + can[y] ) ; }

signed main (int argc , char * argv[]) {
    rep ( i , 1 , 4 ) { can[i] = rint () ; sum += can[i] ; }
    if ( sum & 1 ) return puts ("NO") , 0 ;
    rep ( i , 1 , 4 )
        if ( can[i] == sum - can[i] ) return puts ("YES") , 0 ;
    rep ( i , 1 , 4 ) rep ( j , i + 1 , 4 )
        if ( check ( i , j ) ) return puts ("YES") , 0 ;
    puts ("NO") ;
    return 0 ;
}

CodeForces1230B
题目并不难,直接贪心显然是对的.
肯定是从高位向低位贪心.
因为题目限制不能有前导零,所以首位不能填(0),只能填(1).后面的全赋为(0)即可.
只需要注意,不能有前导零,也就是首位不能填(0),已经是(0)的位置不需要耗费修改次数.

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <string>
#include <vector>
#include <queue>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
#define MEM(x,y) memset ( x , y , sizeof ( x ) )
#define rep(i,a,b) for (int i = (a) ; i <= (b) ; ++ i)
#define per(i,a,b) for (int i = (a) ; i >= (b) ; -- i)
#define pii pair < int , int >
#define one first
#define two second
#define rint read<int>
#define int long long
#define pb push_back
#define db double

using std::queue ;
using std::set ;
using std::pair ;
using std::max ;
using std::min ;
using std::priority_queue ;
using std::vector ;
using std::swap ;
using std::sort ;
using std::unique ;
using std::greater ;

template < class T >
    inline T read () {
        T x = 0 , f = 1 ; char ch = getchar () ;
        while ( ch < '0' || ch > '9' ) {
            if ( ch == '-' ) f = - 1 ;
            ch = getchar () ;
        }
       while ( ch >= '0' && ch <= '9' ) {
            x = ( x << 3 ) + ( x << 1 ) + ( ch - 48 ) ;
            ch = getchar () ;
       }
   return f * x ;
}

const int N = 2e5 + 100 ;

int n , k ;
char s[N] ;
char ans[N] ;

signed main (int argc , char * argv[]) {
    n = rint () ; k = rint () ;
    scanf ("%s" , s + 1 ) ;
    if ( n == 1 ) {
        if ( k ) puts ("0") ;
        else putchar ( s[1] ) ;
        return 0 ;
    }
    int now = 2 ; ans[1] = s[1] ;
    if ( k && s[1] != '1' ) { ans[1] = '1' , -- k ; }
    while ( s[now] == '0' && now <= n ) { ans[now] = s[now] ; ++ now ; }
    for (int i = now ; k && i <= n ; ++ i) {
        ans[i] = '0' ; ++ now ;
        if ( s[i] != '0' ) -- k ;
    }
    rep ( i , now , n ) ans[i] = s[i] ;
    rep ( i , 1 , n ) putchar ( ans[i] ) ;
    return 0 ;
}

CodeForce1230C
比较吓人的一道题.直接做没有什么好的思路.
但数据范围比较小,所以可以直接(brute).
采用深搜即可.枚举对着每个点的骨牌的点数应该是多少.
对骨牌开一个(bool)标记是否用过,能放就放.
这里有一个小技巧,深搜可以不用传统的递归写法.
可以采用(STL)的(next\_permutation)函数生成对着节点的骨牌点数的全排列,注意最大点数为(6)即可.
生成全排列后就可以直接(Theta(n^2))去统计该状态的答案.取(max)即可.

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <string>
#include <vector>
#include <queue>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
#define MEM(x,y) memset ( x , y , sizeof ( x ) )
#define rep(i,a,b) for (int i = (a) ; i <= (b) ; ++ i)
#define per(i,a,b) for (int i = (a) ; i >= (b) ; -- i)
#define pii pair < int , int >
#define one first
#define two second
#define rint read<int>
#define int long long
#define pb push_back
#define db double

using std::queue ;
using std::set ;
using std::pair ;
using std::max ;
using std::min ;
using std::priority_queue ;
using std::vector ;
using std::swap ;
using std::sort ;
using std::unique ;
using std::greater ;

template < class T >
    inline T read () {
        T x = 0 , f = 1 ; char ch = getchar () ;
        while ( ch < '0' || ch > '9' ) {
            if ( ch == '-' ) f = - 1 ;
            ch = getchar () ;
        }
       while ( ch >= '0' && ch <= '9' ) {
            x = ( x << 3 ) + ( x << 1 ) + ( ch - 48 ) ;
            ch = getchar () ;
       }
   return f * x ;
}

const int N = 5e5 + 100 ;

int n , m , ans = - 1 , col[N] ;
bool vis[10][10] , G[10][10] ;

inline int get_it () {
    int res = 0 ; MEM ( vis , 0 ) ;
    rep ( i , 1 , n ) rep ( j , 1 , n )
        if ( G[i][j] ) {
            int u = col[i] , v = col[j] ;
            if ( ! vis[u][v] && ! vis[v][u] )
                ++ res , vis[u][v] = vis[v][u] = true ;
        }
    return res ;
}

signed main (int argc , char * argv[]) {
    n = rint () ; m = rint () ;
    rep ( i , 1 , m ) {
        int u = rint () , v = rint () ;
        G[u][v] = G[v][u] = true ;
    }
    rep ( i , 1 , n ) col[i] = i ; col[7] = 6 ;
    do {
        ans = max ( ans , get_it () ) ;
    } while ( std::next_permutation ( col + 1 , col + n + 1 ) ) ;
    printf ("%lld
" , ans ) ;
    return 0 ;
}

CodeForce1230D
首先,显然每个学生如果在答案集合中,则必定存在另一个学生掌握他所掌握的所有技能.
其次,如果有若干个学生的技能树相同那么他们一定可以同时存在,且缺少其中的某一个学生一定会使答案变劣.
因为所有学生的能力值均为非负整数.
假设我们已经把所有技能树相同的学生都放在了一起,这一定是个合法的答案.
然后我们考虑一个学生(i),如果(i)的技能树是已经在答案中的任意一个学生的真子集(一定是真子集,因为相等都已经在答案里了),那么他就可以安全地进入答案集合中.
否则,要么只选这个学生(i),要么取原集合,取(max)即可.

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <string>
#include <vector>
#include <queue>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
#define MEM(x,y) memset ( x , y , sizeof ( x ) )
#define rep(i,a,b) for (int i = (a) ; i <= (b) ; ++ i)
#define per(i,a,b) for (int i = (a) ; i >= (b) ; -- i)
#define pii pair < int , int >
#define one first
#define two second
#define rint read<int>
#define int long long
#define pb push_back
#define db double

using std::queue ;
using std::set ;
using std::pair ;
using std::max ;
using std::min ;
using std::priority_queue ;
using std::vector ;
using std::swap ;
using std::sort ;
using std::unique ;
using std::greater ;
using std::map ;

template < class T >
    inline T read () {
        T x = 0 , f = 1 ; char ch = getchar () ;
        while ( ch < '0' || ch > '9' ) {
            if ( ch == '-' ) f = - 1 ;
            ch = getchar () ;
        }
        while ( ch >= '0' && ch <= '9' ) {
            x = ( x << 3 ) + ( x << 1 ) + ( ch - 48 ) ;
            ch = getchar () ;
        }
   return f * x ;
}

const int N = 5e5 + 100 ;

int n , ans ; pii s[N] ;
map < int , int > cnt ;
set < int > v ;

signed main (int argc , char * argv[]) {
    n = rint () ;
    rep ( i , 1 , n ) { s[i].one = rint () ; ++ cnt[s[i].one] ; }
    rep ( i , 1 , n ) s[i].two = rint () ;
    for (map < int , int > :: iterator it = cnt.begin () ; it != cnt.end () ; ++ it) {
        if ( it->two <= 1 ) continue ;
        rep ( i , 1 , n ) if ( ( s[i].one | it->one ) == it->one ) v.insert ( i ) ;
    }
    for (set < int > :: iterator it = v.begin () ; it != v.end () ; ++ it)
        ans += s[*it].two ;
    printf ("%lld
" , ans ) ;
    return 0 ;
}