解题分析：

设电梯停在i层，此时的不满意度为Ci

则Ci = ∑_(j=1)^(i-1)▒〖[(b(i-j)*k_(j )+ k_j*1/2*(i-j)(i-j-1)〗]
     + ∑_(j=i+1)^m▒〖[a(j-i)*k_(j )+ k_(j )*1/2*(j-i)(j-i-1)]〗
   C(i+1) = ∑_(j=1)^i▒〖[(b(i+1-j)*k_(j )+ k_j*1/2*(i+1-j)(i-j)〗]
     + ∑_(j=i+2)^m▒〖[a(j-i-1)*k_(j )+ k_(j )*1/2*(j-i-1)(j-i-2)]〗

推出： C(i+1) – Ci = ∑_(j=1)^i▒〖b*〗 k_(j ) + ∑_(j=1)^(i-1)▒〖(i-j)*k_(j ) 〗 + ∑_(j=i+1)^m▒〖(1-a)k_(j ) 〗 - ∑_(j=i+1)^m▒〖(j-i)k_(j ) 〗

令 T[i] = ∑_(j=1)^i▒〖b*〗 k_(j ),H[i] = ∑_(j=1)^(i-1)▒〖(i-j)*k_(j ) 〗,M[i] = ∑_(j=i+1)^m▒〖(1-a)k_(j ) 〗,S[i] = ∑_(j=i+1)^m▒〖(j-i)k_(j ) 〗
则T[i+1] = T[i] + b*k_(i+1 ) T[1] = b*k_(1 )
H[i+1] = ∑_(j=1)^i▒〖(i+1-j)*k_(j ) 〗 = ∑_(j=1)^(i-1)▒〖(i-j+1)*k_(j ) 〗 + k_(i )=H[i] + ∑_(j=1)^i▒k_(j ) H[1] = 0;
M[i+1] =  ∑_(j=i+2)^m▒〖(1-a)k_(j ) 〗 = ∑_(j=i+1)^m▒〖(1-a)k_(j ) 〗 - (1-a) k_(i+1 ) = M[i] -(1-a) k_(i+1 ) 
M[i] = M[i+1] + (1-a) k_(i+1 ),M[m] = 0;
S[i+1] =  ∑_(j=i+2)^m▒〖(j-i-1)k_(j ) 〗 =  ∑_(j=i+1)^m▒〖(j-i-1)k_(j ) 〗 = S[i] –∑_(j=i+1)^m▒k_(j ) ;
S[i] = S[i+1] +∑_(j=i+1)^m▒k_(j ) ,S[m] = 0;

令 sum[i] = j=1ik_(j ), Sum = sum[m]; sum[1] = k_(1 )
则 H[i+1] = H[i]+sum[i]; S[i] = S[i+1] + Sum-sum[i];

C1 按式子先算出来。。
则C(i+1) = Ci + T[i] + H[i] + M[i] – S[i];

求出数组Ci的最小值即可。

• Re:公式数学推导 (1233) CSUST_14 2013-04-20 11:05:56