Codeforces round #128 (Div. 2)

Reprint please indicate the source, thank you http://blog.csdn.net/ACM_cxlove? Viewmode = contents by --- cxlove

A: Two Problems

Many people have planted very difficult questions. There are two questions: the initial score is A, the score is reduced by da per minute, and the initial score is B, and the database is reduced per minute. Ask if a person just gets x points.

X may be 0, which makes wa a lot of people, and x 0. This is certainly acceptable, and neither of the two questions is answered.

After that, the O (t) practice continued to be wa, speechless, and forced to directly attack O (T * t.

# Include <iostream> # include <iomanip> # include <cstdio> # include <cstring> # include <algorithm> # include <cstdlib> # include <cmath> # include <map> # include <stack> # include <set> # include <queue> # include <string> # include <vector> # define EPS 1e-6 # define ll long # define LD long double # define PI ACOs (-1.0) # define INF 1ll <60 using namespace STD; int flag [605]; int main () {int x, t, A, da, B, DB, I, j, AA, BB; scanf ("% d", & X, & T, & A, & B, & Da, & dB ); if (x = 0) puts ("yes"); else {for (I = 0; I <t; I ++) for (j = 0; j <t; j ++) {AA = A-I * Da; BB = B-J * dB; if (AA = x | BB = x | AA + BB = x) {puts ("yes"); Return 0 ;}} puts ("no");} return 0 ;}
 

B: Game on paper

It turns out that there were no evaluations in the end. The practice is to place a piece of chess, with the 9-palace lattice centered on the piece as the center to determine whether a 3*3 is formed. The maximum number of pawns is 10 W, and the maximum number of pawns is 10 W * 9*9. Note that the enumeration in main is in nine positions, which are held here at that time.

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                # define EPS 1e-6 # define ll long # define LD long double # define PI ACOs (-1.0) # define INF 1ll <60 using namespace STD; bool flag [1005] [1005]; int way [9] [2] ={{}, {-1, 0 },{}, {0,-1}, {}, {1,-1}, {-}, {-1,-1 }, {0, 0 }}; bool check (int x, int y) {If (flag [x] [Y] = false) return false; For (INT I = 0; I <8; I ++) if (flag [x + way [I] [0] [Y + way [I] [1] = false) return false; return true;} int main () {int n, m; while (scanf ("% d", & N, & M )! = EOF) {memset (flag, false, sizeof (FLAG); int X, Y; int ans =-1; for (int t = 1; t <= m; t ++) {scanf ("% d", & X, & Y); flag [x] [Y] = true; int I; If (ANS! =-1) continue; for (I = 0; I <9; I ++) {If (check (x + way [I] [0], Y + way [I] [1]) {ans = T; break ;}} printf ("% d \ n", ANS) ;}return 0 ;} 
              
             
            
           
          
        
       
      
     
    
   
  

 

C: photographer

Relatively simple and greedy. Customers with relatively small requirements can be satisfied first, and sorted in ascending order of total consumption.

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                # define EPS 1e-6 # define ll long # define LD long double # define PI ACOs (-1.0) # define INF 1ll <60 using namespace STD; int n, a, B, d; struct node {int X, Y; int cost; int ID X;} A [100005]; bool CMP (node N1, node N2) {return n1.cost 
               
                 = A [I]. cost) {ans [CNT ++] = A [I]. idx; D-= A [I]. cost;} elsebreak; printf ("% d \ n", CNT); If (CNT! = 0) {for (INT I = 0; I 
                
               
              
             
            
           
          
        
       
      
     
    
   
  

D: hit Ball

A messy explanation of the question. In a corridor, standing in (A/2, M, 0) without a vector (VX, Vy, VZ) throws a ball and asks where the ball finally hits the door. The door width is A, the door height is B, there is a wall on both sides of the corridor, there is a ceiling on the upper side, it will be reflected after the impact.

First, according to the y-axis direction, the component of the speed on the y-axis is always Vy, and is not changed by the impact. You can calculate the time-M/Vy;

Since each reflection does not change the speed component, it only changes the direction. For example, when it hits a wall and the speed changes from vx to-VX, the distance between the X axis direction motion can be calculated based on the known time, and then fold it into the range 0-.

# Include <iostream> # include <iomanip> # include <cstdio> # include <cstring> # include <algorithm> # include <cstdlib> # include <cmath> # include <map> # include <stack> # include <set> # include <queue> # include <string> # include <vector> # define EPS 1e-7 # define ll long # define LD long double # define PI ACOs (-1.0) # define INF 1ll <60 using namespace STD; int A, B, M; int VX, Vy, VZ; int main () {While (scanf ("% d", & A, & B, & M, & VX, & V Y, & VZ )! = EOF) {double T = m * (-1.0)/Vy; Double X = T * VX + A/2.0, ansx; If (x <EPS) {x =-X; int Cx = (INT) (X/a); ansx = x-cx * A; If (CX & 1) ansx = A-ansx ;} else if (x> A + EPS) {X-= A; int Cx = (INT) (X/a); ansx = x-cx * A; If (! (CX & 1) ansx = A-ansx;} elseansx = x; Double Z = T * VZ; int cz = (INT) (z/B ); double ansz = z-CZ * B; If (CZ & 1) ansz = B-ansz; printf ("%. 10f %. 10f \ n ", ansx, ansz);} return 0 ;}
 

E: readforces. You have time to read the questions. However, the English is really poor.