HDU 1729 SG Function

Stone gametime limit: 5000/1000 ms (Java/other) Memory limit: 65535/32768 K (Java/other)Total submission (s): 12 Accepted submission (s): 6Font: Times New Roman | Verdana
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1. There are n boxes; each box has its size. The box can hold up to S stones if the size is S.
2. At the beginning of the game, there are some stones in these boxes.
3. the players take turns choosing a box and put a number of stones into the box. the number mustn't be great than the square of the number of stones before the player adds the stones. for example, the player can add 1 to 9 stones if there are 3 stones in
Box. Of course, the total number of stones mustn't be great than the size of the box.
4. Who can't add stones any more will loss the game.

Give an initial state of the game. You are supposed to find whether the first player will win the game if both of the players make the best strategy. inputthe input file contains several test cases.
Each test case begins with an integer N, 0 <n ≤ 50, the number of the boxes.
In the next N line there are two integer Si, CI (0 ≤ CI ≤ Si ≤1,000,000) on each line, as the size of the box is SI and there are CI stones in the box.
N = 0 indicates the end of input and shoshould not be processed. outputfor each test case, output the number of the case on the first line, then output "yes" (without quotes) on the next line if the first player can win the game, otherwise output "no ". sample Input

3 

2 0 

3 3 

6 2 

2 

6 3 

6 3 

0

Sample output

 Case 1: 

 Yes 

 Case 2: 

 No

This SG has been depressing for me for a long time. I have consulted some people and asked Baidu to check it out ......

First, let's assume that C is a must-have point, then C + C * C <S (even if the largest C * C cannot be filled up)

C + 1 is a winning point. C + 1 + (C + 1) * (C + 1)> = s based on the two formulas, we can find the maximum defeat point t smaller than S.

T + 1 ---- S-1 is a winning point, because (t + 1 --- S-1 can all reach the point S ).... (Principle 1, as long as it can reach the point of defeat, it must be a point of victory, 2, each step can only reach the point of victory, then this must be a point of defeat)

The following three cases are discussed:

 When c = TAt this time, it must be a point of failure and 0 is returned directly; 

When C> T, C is the winning point, as long as the minimum value in the subsequent set of C is returned... The minimum value is s-c.

s  s-1   s-2  s-3 .....  c .. .. t   t-1  t-2  t-3   .................... 1 

0   1  2   3   .....  s-c ...   0 

When C <t, t is used as s to continue recursion .. It should always be a point of defeat for the newly opened S...

My code:

#include<stdio.h> 

#include<math.h> 

int getsg(int s,int c) 

{ 

  int t; 

  t=(int)sqrt(s); 

 while(t*t+t>=s) 

  { 

               t--; 

    } 

       if(c>t) 

          return s-c; 

     else if(c==t) 

           return 0; 

       else 

            return getsg(t,c); 

} 

int main() 

{ 

      int t,x,y,count=1,c,i; 

  while(scanf("%d",&t),t) 

 { 

               c=0; 

            for(i=0;i<t;i++) 

         { 

                       scanf("%d%d",&x,&y); 

                    c^=getsg(x,y); 

          } 

               printf("Case %d:\n",count++); 

           if(c) 

                   printf("Yes\n"); 

                else 

                    printf("No\n"); 

 } 

       return 0; 

}