CodeForces Round #516 Div2 题解

A. Make a triangle!

暴力...

就是给你三个数，你每次可以选一个加1，问最少加多少次能构成三角形

#include <bits/stdc++.h>

#define ll long long
#define inf 0x3f3f3f3f
#define il inline
#define in1(a) read(a)
#define in2(a,b) in1(a),in1(b)
#define in3(a,b,c) in2(a,b),in1(c)
#define in4(a,b,c,d) in2(a,b),in2(c,d)

inline void read( int &x ){
    x =  ; int f =  ; char c = getchar() ;
    while( c < '' || c > '' ) {
        if( c == '-' ) f = - ;
        c = getchar() ;
    }
    while( c >= '' && c <= '' ) {
        x = (x << ) + (x << ) + c -  ;
        c = getchar() ;
    }
    x *= f ;
}

inline void readl( ll &x ){
    x =  ; ll f =  ; char c = getchar() ;
    while( c < '' || c > '' ) {
        if( c == '-' ) f = - ;
        c = getchar() ;
    }
    while( c >= '' && c <= '' ) {
        x = (x << ) + (x << ) + c -  ;
        c = getchar() ;
    }
    x *= f ;
}

using namespace std ;

#define N 100010

int a , b , c ;

int main(){
    in3( a ,b , c ) ;
    if( a > b ) swap( a , b ) ;
    if( b > c ) swap( b , c ) ;
    if( a > c ) swap( a , c ) ;
    int ans =  ;
    while( a + b <= c ) {
        a ++ ;
        ans ++ ;
        if( a > b ) swap( a , b ) ;
    }
    printf( "%d\n" , ans ) ;
    return  ;
}

B. Equations of Mathematical Magic

求满足式子$a-(a\ xor\ x)-x=0$的$x$值的数量

$a<=2^{30}-1$

对于某一位的$a$和$x$

设$a$这一位上的数字为$a_i$，$x$这一位上的数字为$x_i$

对于$a_i=0$，只有$x_i=0$才成立

对于$a_i=1$，$x_i=0$或者$x_i=1$均成立

所以乘法原理乘一下就好了

ZincSabian聚聚抬了一手（orz我B题就不会了

#include <cstdio>
#include <cstring>
#include <algorithm>

#define ll long long
#define inf 0x3f3f3f3f
#define il inline
#define in1(a) readl(a)
#define in2(a,b) in1(a),in1(b)
#define in3(a,b,c) in2(a,b),in1(c)
#define in4(a,b,c,d) in2(a,b),in2(c,d)

inline void read( int &x ){
    x =  ; int f =  ; char c = getchar() ;
    while( c < '' || c > '' ) {
        if( c == '-' ) f = - ;
        c = getchar() ;
    }
    while( c >= '' && c <= '' ) {
        x = (x << ) + (x << ) + c -  ;
        c = getchar() ;
    }
    x *= f ;
}

inline void readl( ll &x ){
    x =  ; ll f =  ; char c = getchar() ;
    while( c < '' || c > '' ) {
        if( c == '-' ) f = - ;
        c = getchar() ;
    }
    while( c >= '' && c <= '' ) {
        x = (x << ) + (x << ) + c -  ;
        c = getchar() ;
    }
    x *= f ;
}

using namespace std ;

#define N 100010

ll T , a ;

int main() {
    in1( T ) ;
    while( T -- ) {
        in1( a ) ;
        ll cnt =  ;
        for( int i =  ; i >=  ; i -- ) {
            if( a&(1ll<<i) ) cnt *= 2ll ;
        }
        printf( "%lld\n" , cnt ) ;
    }
}

C. Oh Those Palindromes

怎么B,C都是结论题啊

C题直接把所有一样的数排在一起就好了...

因为这样确实就能排出最大的回文串了

可以找几个字符串自己写写画画一下，感性理解

#include <cstdio>
#include <cstring>
#include <algorithm>

#define ll long long
#define inf 0x3f3f3f3f
#define il inline
#define in1(a) read(a)
#define in2(a,b) in1(a),in1(b)
#define in3(a,b,c) in2(a,b),in1(c)
#define in4(a,b,c,d) in2(a,b),in2(c,d)

inline void read( int &x ){
    x =  ; int f =  ; char c = getchar() ;
    while( c < '' || c > '' ) {
        if( c == '-' ) f = - ;
        c = getchar() ;
    }
    while( c >= '' && c <= '' ) {
        x = (x << ) + (x << ) + c -  ;
        c = getchar() ;
    }
    x *= f ;
}

inline void readl( ll &x ){
    x =  ; ll f =  ; char c = getchar() ;
    while( c < '' || c > '' ) {
        if( c == '-' ) f = - ;
        c = getchar() ;
    }
    while( c >= '' && c <= '' ) {
        x = (x << ) + (x << ) + c -  ;
        c = getchar() ;
    }
    x *= f ;
}

using namespace std ;

#define N 100010

int n ;
char ch[ N ] ;

int main(){
    in1( n ) ;
    scanf( "%s" , ch+ ) ;
    sort( ch+ , ch+n+ ) ;
    printf( "%s" , ch+ ) ;
}

D. Labyrinth

赛时没调出来...

然后赛后3min调出来

$system\ test$ 完后交了就过了

哭...掉分了

唔...其实这题就是建两个图，对于第一个图只有向左走才会花费代价，对于第二个图只有向右走才会花费代价

然后就是码码码了，我写了$4KB$...

写的$spfa$，网格图居然没卡$spfa$...

代码可能有点丑...

将就着看吧

#include <cstdio>
#include <cstring>
#include <algorithm>

#define ll long long
#define debug printf("233\n")
#define inf 0x3f3f3f3f
#define il inline
#define in1(a) read(a)
#define in2(a,b) in1(a),in1(b)
#define in3(a,b,c) in2(a,b),in1(c)
#define in4(a,b,c,d) in2(a,b),in2(c,d)

inline void read( int &x ){
    x =  ; int f =  ; char c = getchar() ;
    while( c < '' || c > '' ) {
        if( c == '-' ) f = - ;
        c = getchar() ;
    }
    while( c >= '' && c <= '' ) {
        x = (x << ) + (x << ) + c -  ;
        c = getchar() ;
    }
    x *= f ;
}

inline void readl( ll &x ){
    x =  ; ll f =  ; char c = getchar() ;
    while( c < '' || c > '' ) {
        if( c == '-' ) f = - ;
        c = getchar() ;
    }
    while( c >= '' && c <= '' ) {
        x = (x << ) + (x << ) + c -  ;
        c = getchar() ;
    }
    x *= f ;
}

using namespace std ;

#define N 2010

int n , m , r , c , x , y ;
char ch[ N ][ N ] ;
struct edge {
    int to , nxt , v ;
} e[ N * N *  ] , E[ N * N *  ];
int head[ N * N ] , cnt , Head[ N * N ] , Cnt ;
int vis[ N * N ] , d[ N * N ] , Vis[ N * N ] , D[ N * N ] ;
int q[  ] ;

void ins1( int u , int v , int w ) {
    e[ ++ cnt ].to = v ;
    e[ cnt ].nxt = head[ u ] ;
    e[ cnt ].v = w ;
    head[ u ] = cnt ;
}

void ins2( int u , int v , int w ) {
    E[ ++ Cnt ].to = v ;
    E[ Cnt ].nxt = Head[ u ] ;
    E[ Cnt ].v = w ;
    Head[ u ] = Cnt ;
}

void spfa() {
    int s = (r-) * m + c ;
    q[  ] = s ;
    int l =  , r =  ;
    for( int i =  ; i <= n * m ; i ++ ) d[ i ] = inf ;
    d[ s ] =  ;
    vis[ s ] =  ;
    while( l != r ) {
        int u = q[ l ++ ] ;
        vis[ u ] =  ;
        if( l ==  ) l =  ;
        for( int i = head[ u ] ; i ;  i = e[ i ].nxt ) {
            int v = e[ i ].to ;
            if( d[ v ] > d[ u ] + e[ i ].v ) {
                d[ v ] = d[ u ] + e[ i ].v ;
                if( !vis[ v ] ) {
                    vis[ v ] =  ;
                    q[ r ++ ] = v ;
                    if( r ==  ) r =  ;
                }
            }
        }
    }
}

void spfa2() {
    int s = (r-) * m + c ;
    int l =  , r =  ;
    q[  ] = s ;Vis[ s ] =  ;
    for( int i =  ; i <= n * m ; i ++ ) D[ i ] = inf ;
    D[ s ] =  ;
    while( l != r ) {
        int u = q[ l ++ ] ;
        Vis[ u ] =  ;
        if( l ==  ) l =  ;
        for( int i = Head[ u ] ; i ;  i = E[ i ].nxt ) {
            int v = E[ i ].to ;
            if( D[ v ] > D[ u ] + E[ i ].v ) {
                D[ v ] = D[ u ] + E[ i ].v ;
                if( !Vis[ v ] ) {
                    Vis[ v ] =  ;
                    q[ r ++ ] = v ;
                    if( r ==  ) r =  ;
                }
            }
        }
    }
}

int main(){
    in2( n , m ) ;
    in2( r , c ) ;
    in2( x , y ) ;
    for( int i =  ; i <= n ; i ++ ) {
        scanf( "%s" , ch[ i ] +  ) ;
    }
    for( int i =  ; i <= n ; i ++ ) {
        for( int j =  ; j <= m ; j ++ ) {
            if( ch[ i ][ j ] == '*' ) continue ;
            if(i->=&&ch[i-][j]=='.') ins1((i-)*m+j,(i-)*m+j,),  ins2((i-)*m+j,(i-)*m+j,);
            if(j->=&&ch[i][j-]=='.') ins1((i-)*m+j,(i-)*m+j-,),ins2((i-)*m+j,(i-)*m+j-,);
            if(i+<=n&&ch[i+][j]=='.') ins1((i-)*m+j,i*m+j,),      ins2((i-)*m+j,i*m+j,);
            if(j+<=m&&ch[i][j+]=='.') ins1((i-)*m+j,(i-)*m+j+,),ins2((i-)*m+j,(i-)*m+j+,);
        }
    }
    spfa() ;
    spfa2() ;
    int ans =  ;
    for( int i =  ; i <= n * m ; i ++ ) {
        if( d[ i ] <= x && D[ i ] <= y ) ans ++ ;
    }
    printf( "%d\n" , ans ) ;
}

E. Dwarves, Hats and Extrasensory Abilities

这题有点思路，晚上补一下

---

因为题目保证了白的在一边黑的在一边

所以我们可以二分这个分界点

然后输出的时候，纵坐标随便找两个，尽量让他们连线与坐标轴成45度角那样子

反正不要太歪，我取的纵坐标是$0$和$2$

就是真的很不习惯交互的形式

#include <bits/stdc++.h>

#define ll long long
#define inf 0x3f3f3f3f
#define il inline
#define in1(a) readl(a)

inline void readl( ll &x ){
    x =  ; ll f =  ; char c = getchar() ;
    while( c < '' || c > '' ) {
        if( c == '-' ) f = - ;
        c = getchar() ;
    }
    while( c >= '' && c <= '' ) {
        x = (x << ) + (x << ) + c -  ;
        c = getchar() ;
    }
    x *= f ;
}

using namespace std ;

#define N 100010

ll n ;
ll x , y ;
char s[ N ] ;

int main() {
    in1( n ) ;
    puts( "0 0" ) ;
    fflush( stdout ) ;
    scanf( "%s" , s ) ;
    char tmp = s[  ] ;
    ll l =  , r = 1e9 ;
    for( int i =  ; i <= n ; i ++ ) {
        ll mid = ( l + r ) >>  ;
        printf( "%lld %lld\n" , mid , 1ll ) ;
        fflush( stdout ) ;
        scanf( "%s" , s ) ;
        if( s[  ] == tmp ) l = mid +  ;
        else r = mid -  ;
    }
    printf( "%lld 0 %lld 2\n" , l , r ) ;
}

F. Candies for Children

以后补...