HDU 2085 核反应堆

此题找出关系即可

设n微秒时a( n ) 为高能粒子个数，b ( n )为低能粒子个数;

  经分析可得 a ( n ) = 3 * a( n - 1 ) + 2 * b ( n - 1 ), b ( n ) = a ( n - 1 ) + b ( n - 1 );

然后直接打表即可，还有要注意要用long long 型存储

#include <stdio.h>
long long ch[ 35 ][ 2 ];
void chart (  )
{
     long long a,b,c;//a表示高能粒子，b表示低能粒子，c记录上次的高能粒子
     int i = 1;
     ch[0][0] = 1;
     ch[0][1] = 0;
     for ( a = 1 , b = 0 , c = 1; i < 35 ; ++i )
     {
         a = 3 * a + 2 * b;
         b = c + b;
         c = a;
         ch[ i ][ 0 ] = a;
         ch[ i ][ 1 ] = b;
     } 
 }
int main ( )
{
    chart(  );
    int n;
    while ( scanf ( "%d" , &n ) == 1 && n != -1 )
          printf ( "%I64d, %I64d\n" , ch[n][0],ch[n][1] );
    return 0;
}