hdu2046 骨牌铺方格
题解：
递推题，如图，
代码如下：
#include <stdio.h>__int64 A[51][30], B[51];int main(){int n, i, j;for (i=0; i<=50; i++)A[i][0] = 1;for (i=2; i<=50; i++){for (j=1; j<i/2+1; j++)A[i][j] = A[i-2][j-1] + A[i-1][j];}for (i=1; i<=50; i++){for (j=0; j<30; j++)B[i] += A[i][j];}while (scanf("%d", &n) != EOF)printf("%I64d\n", B[n]);return 0;}
如果仔细观察的话，可以发现递推式：A[i] = A[i-2] + A[i-1];
代码如下：
#include <stdio.h>__int64 A[51] = {0, 1, 2, 3};int main(){int n, i;for (i=3; i<=50; i++)A[i] = A[i-2] + A[i-1];while (scanf("%d", &n) != EOF)printf("%I64d\n", A[n]);return 0;}