zoj 2771 - Get Out of the Glass
2014-09-18 21:55
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题目:有三层玻璃叠在一起,一束斜着照射的光纤从最上面射入,问有多少条光线反射n次。
分析:dp,分成奇偶两种状况考虑;
奇数情况:每个面的光线等于上次的本平面以上的点的反射;
偶数情况:每个面的光线等于上次的本平面一下的点的反射。
说明:(2011-09-19 01:29)。
分析:dp,分成奇偶两种状况考虑;
奇数情况:每个面的光线等于上次的本平面以上的点的反射;
偶数情况:每个面的光线等于上次的本平面一下的点的反射。
说明:(2011-09-19 01:29)。
#include <stdio.h> #include <string.h> long long F[ 61 ][ 4 ]; int main() { int i,j,k,n; memset( F, 0L, sizeof( F ) ); F[ 0 ][ 0 ] = 1L; for ( i = 1 ; i <= 60 ; ++ i ) for ( j = 0 ; j <= 3 ; ++ j ) if ( i%2 ) { for ( k = 0 ; k < j ; ++ k ) F[ i ][ j ] += F[ i-1 ][ k ]; }else { for ( k = j+1 ; k <= 3 ; ++ k ) F[ i ][ j ] += F[ i-1 ][ k ]; } while ( scanf("%d",&n) != EOF ) { long long sum = 0; for ( i = 0 ; i < 3 ; ++ i ) sum += F[ n ][ i+n%2 ]; printf("%lld\n",sum); } return 0; }
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