学步园

传送门：Codeforces Round #276 (Div. 2)

A.485A Factory

按照题目规则改变a，a大于m时没意义，所以对m取模，根据题意，当a变化时变回到以前出现过的a时，说明出现循环，无解，否则会一直进行到出现a%m==0为止。

#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;

typedef long long LL ;

#define rep( i , a , b ) for ( int i = ( a ) ; i <  ( b ) ; ++ i )
#define For( i , a , b ) for ( int i = ( a ) ; i <= ( b ) ; ++ i )
#define rev( i , a , b ) for ( int i = ( a ) ; i >= ( b ) ; -- i )
#define travel( e , H , u ) for ( Edge* e = H[u] ; e ; e = e -> next )
#define clr( a , x ) memset ( a , x , sizeof a )

const int MAXN = 100005 ;

int vis[MAXN] ;
int a , m ;

void solve () {
    clr ( vis , 0 ) ;
    int ok = 0 ;
    while ( 1 ) {
        if ( a % m == 0 ) {
            ok = 1 ;
            break ;
        } else if ( vis[a % m] ) break ;
        vis[a % m] = 1 ;
        a = ( a + a % m ) % m ;
    }
    printf ( ok ? "Yes\n" : "No\n" ) ;
}

int main () {
    while ( ~scanf ( "%d%d" , &a , &m ) ) solve () ;
    return 0 ;
}

B.485B Valuable Resources

ans = max（maxx - minx ， maxy - miny ）^ 2。

#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;

typedef long long LL ;

#define rep( i , a , b ) for ( int i = ( a ) ; i <  ( b ) ; ++ i )
#define For( i , a , b ) for ( int i = ( a ) ; i <= ( b ) ; ++ i )
#define rev( i , a , b ) for ( int i = ( a ) ; i >= ( b ) ; -- i )
#define travel( e , H , u ) for ( Edge* e = H[u] ; e ; e = e -> next )
#define clr( a , x ) memset ( a , x , sizeof a )

const int INF = 0x3f3f3f3f ;

int n ;

void solve () {
    LL minx = INF , miny = INF ;
    LL maxx = -INF , maxy = -INF ;
    LL x , y ;
    For ( i , 1 , n ) {
        scanf ( "%I64d%I64d" , &x , &y ) ;
        minx = min ( minx , x ) ;
        maxx = max ( maxx , x ) ;
        miny = min ( miny , y ) ;
        maxy = max ( maxy , y ) ;
    }
    LL ans = max ( maxx - minx , maxy - miny ) ;
    printf ( "%I64d\n" , ans * ans ) ;
}

int main () {
    while ( ~scanf ( "%d" , &n ) ) solve () ;
    return 0 ;
}

C.485C Bits

从l的最低位开始每位挨个填1直到l>r为止，这样一定是1个数最多的且最小的，比这个1多的肯定不存在，剩下的都是比他大的。

#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;

typedef long long LL ;

#define rep( i , a , b ) for ( int i = ( a ) ; i <  ( b ) ; ++ i )
#define For( i , a , b ) for ( int i = ( a ) ; i <= ( b ) ; ++ i )
#define rev( i , a , b ) for ( int i = ( a ) ; i >= ( b ) ; -- i )
#define travel( e , H , u ) for ( Edge* e = H[u] ; e ; e = e -> next )
#define clr( a , x ) memset ( a , x , sizeof a )

LL l , r ;
int n ;

void solve () {
    scanf ( "%I64d%I64d" , &l , &r ) ;
    //LL ans = 0 ;
    rep ( i , 0 , 63 ) {
        if ( ( l | ( 1LL << i ) ) <= r ) {
            l |= ( 1LL << i ) ;
        }
    }
    printf ( "%I64d\n" , l ) ;
}

int main () {
    int T ;
    scanf ( "%d" , &T ) ;
    while ( T -- ) solve () ;
    return 0 ;
}

D.485D Maximum Value

我们可以对每个数ai的所有倍数，找从左端靠近的最接近这个倍数的数aj，同时要满足aj>ai，然后更新ans = max （ans，aj%ai）。

我们可以预处理出1~2000000内每个数x从左端靠近的最接近这个数的数，类似递推：

nearest[i] = hash[i - 1] ? i - 1 : nearest[i - 1]。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;

#define rep( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define For( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define rev( i , a , b ) for ( int i = a ; i >= b ; -- i )
#define clr( a , x ) memset ( a , x , sizeof a )

const int MAXN = 1000005 ;

int nearest[MAXN << 1] ;
bool hash[MAXN << 1] ;
int n ;

void solve () {
	int x , ans = 0 ;
	clr ( hash , 0 ) ;
	clr ( nearest , 0 ) ;
	For ( i , 1 , n ) {
		scanf ( "%d" , &x ) ;
		hash[x] = 1 ;
	}
	For ( i , 1 , 2000000 ) {
		if ( hash[i - 1] ) nearest[i] = i - 1 ;
		else nearest[i] = nearest[i - 1] ;
	}
	For ( i , 2 , 1000000 ) if ( hash[i] && ans < i - 1 ) {
		for ( int j = i + i ; j < 1000000 + i ; j += i ) {
			if ( nearest[j] > i ) {
				ans = max ( ans , nearest[j] % i ) ;
				if ( ans == i - 1 ) break ;
			}
		}
	}
	printf ( "%d\n" , ans ) ;
}

int main () {
	while ( ~scanf ( "%d" , &n ) ) solve () ;
	return 0 ;
}