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 = 0 ; int f = 1 ; char c = getchar() ; while( c < '0' || c > '9' ) { if( c == '-' ) f = -1 ; c = getchar() ; } while( c >= '0' && c <= '9' ) { x = (x << 1) + (x << 3) + c - 48 ; c = getchar() ; } x *= f ; } inline void readl( ll &x ){ x = 0 ; ll f = 1 ; char c = getchar() ; while( c < '0' || c > '9' ) { if( c == '-' ) f = -1 ; c = getchar() ; } while( c >= '0' && c <= '9' ) { x = (x << 1) + (x << 3) + c - 48 ; 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 = 0 ; while( a + b <= c ) { a ++ ; ans ++ ; if( a > b ) swap( a , b ) ; } printf( "%d " , ans ) ; return 0; }
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 = 0 ; int f = 1 ; char c = getchar() ; while( c < '0' || c > '9' ) { if( c == '-' ) f = -1 ; c = getchar() ; } while( c >= '0' && c <= '9' ) { x = (x << 1) + (x << 3) + c - 48 ; c = getchar() ; } x *= f ; } inline void readl( ll &x ){ x = 0 ; ll f = 1 ; char c = getchar() ; while( c < '0' || c > '9' ) { if( c == '-' ) f = -1 ; c = getchar() ; } while( c >= '0' && c <= '9' ) { x = (x << 1) + (x << 3) + c - 48 ; c = getchar() ; } x *= f ; } using namespace std ; #define N 100010 ll T , a ; int main() { in1( T ) ; while( T -- ) { in1( a ) ; ll cnt = 1 ; for( int i = 40 ; i >= 0 ; i -- ) { if( a&(1ll<<i) ) cnt *= 2ll ; } printf( "%lld " , 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 = 0 ; int f = 1 ; char c = getchar() ; while( c < '0' || c > '9' ) { if( c == '-' ) f = -1 ; c = getchar() ; } while( c >= '0' && c <= '9' ) { x = (x << 1) + (x << 3) + c - 48 ; c = getchar() ; } x *= f ; } inline void readl( ll &x ){ x = 0 ; ll f = 1 ; char c = getchar() ; while( c < '0' || c > '9' ) { if( c == '-' ) f = -1 ; c = getchar() ; } while( c >= '0' && c <= '9' ) { x = (x << 1) + (x << 3) + c - 48 ; c = getchar() ; } x *= f ; } using namespace std ; #define N 100010 int n ; char ch[ N ] ; int main(){ in1( n ) ; scanf( "%s" , ch+1 ) ; sort( ch+1 , ch+n+1 ) ; printf( "%s" , ch+1 ) ; }
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 ") #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 = 0 ; int f = 1 ; char c = getchar() ; while( c < '0' || c > '9' ) { if( c == '-' ) f = -1 ; c = getchar() ; } while( c >= '0' && c <= '9' ) { x = (x << 1) + (x << 3) + c - 48 ; c = getchar() ; } x *= f ; } inline void readl( ll &x ){ x = 0 ; ll f = 1 ; char c = getchar() ; while( c < '0' || c > '9' ) { if( c == '-' ) f = -1 ; c = getchar() ; } while( c >= '0' && c <= '9' ) { x = (x << 1) + (x << 3) + c - 48 ; 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 * 4 ] , E[ N * N * 4 ]; int head[ N * N ] , cnt , Head[ N * N ] , Cnt ; int vis[ N * N ] , d[ N * N ] , Vis[ N * N ] , D[ N * N ] ; int q[ 1000100 ] ; 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-1) * m + c ; q[ 1 ] = s ; int l = 1 , r = 2 ; for( int i = 1 ; i <= n * m ; i ++ ) d[ i ] = inf ; d[ s ] = 0 ; vis[ s ] = 1 ; while( l != r ) { int u = q[ l ++ ] ; vis[ u ] = 0 ; if( l == 1000000 ) l = 1 ; 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 ] = 1 ; q[ r ++ ] = v ; if( r == 1000000 ) r = 1 ; } } } } } void spfa2() { int s = (r-1) * m + c ; int l = 1 , r = 2 ; q[ 1 ] = s ;Vis[ s ] = 1 ; for( int i = 1 ; i <= n * m ; i ++ ) D[ i ] = inf ; D[ s ] = 0 ; while( l != r ) { int u = q[ l ++ ] ; Vis[ u ] = 0 ; if( l == 1000000 ) l = 1 ; 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 ] = 1 ; q[ r ++ ] = v ; if( r == 1000000 ) r = 1 ; } } } } } int main(){ in2( n , m ) ; in2( r , c ) ; in2( x , y ) ; for( int i = 1 ; i <= n ; i ++ ) { scanf( "%s" , ch[ i ] + 1 ) ; } for( int i = 1 ; i <= n ; i ++ ) { for( int j = 1 ; j <= m ; j ++ ) { if( ch[ i ][ j ] == '*' ) continue ; if(i-1>=1&&ch[i-1][j]=='.') ins1((i-1)*m+j,(i-2)*m+j,0), ins2((i-1)*m+j,(i-2)*m+j,0); if(j-1>=1&&ch[i][j-1]=='.') ins1((i-1)*m+j,(i-1)*m+j-1,1),ins2((i-1)*m+j,(i-1)*m+j-1,0); if(i+1<=n&&ch[i+1][j]=='.') ins1((i-1)*m+j,i*m+j,0), ins2((i-1)*m+j,i*m+j,0); if(j+1<=m&&ch[i][j+1]=='.') ins1((i-1)*m+j,(i-1)*m+j+1,0),ins2((i-1)*m+j,(i-1)*m+j+1,1); } } spfa() ; spfa2() ; int ans = 0 ; for( int i = 1 ; i <= n * m ; i ++ ) { if( d[ i ] <= x && D[ i ] <= y ) ans ++ ; } printf( "%d " , 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 = 0 ; ll f = 1 ; char c = getchar() ; while( c < '0' || c > '9' ) { if( c == '-' ) f = -1 ; c = getchar() ; } while( c >= '0' && c <= '9' ) { x = (x << 1) + (x << 3) + c - 48 ; 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[ 0 ] ; ll l = 1 , r = 1e9 ; for( int i = 2 ; i <= n ; i ++ ) { ll mid = ( l + r ) >> 1 ; printf( "%lld %lld " , mid , 1ll ) ; fflush( stdout ) ; scanf( "%s" , s ) ; if( s[ 0 ] == tmp ) l = mid + 1 ; else r = mid - 1 ; } printf( "%lld 0 %lld 2 " , l , r ) ; }
F. Candies for Children
以后补...