借人气问一下CHANEL的钱包什么价格呀 - 未名空间MITBBS历史存档

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借人气问一下CHANEL的钱包什么价格呀

借人气问一下CHANEL的钱包什么价格呀# Fashion - 美丽时尚

s*x2011-07-29 07:07

1 楼

网上找了半天也没找到。多谢！

s*u2011-07-29 07:07

2 楼

Several manuscripts will be submitted to building energy related journals (
IF from 2 to 5) soon and suggested reviewers are needed.
If you are interested, please send me your name, institution, email address
and a short description of yourself.
Thanks a lot!

N*n2011-07-29 07:07

3 楼

It's been almost 30 months since Silver last visited 20/oz area. Silver
bears are trying hard to drive it down now.

S*12011-07-29 07:07

4 楼

朋友让我帮忙买香奈儿的钱包，香包不是只有店里才买得到的嘛，我也不想专门帮她跑
一趟看价格，有哪位好心的MM告知一下哈。给个价格范围也行~~谢谢啦啦啦~~

k*r2011-07-29 07:07

5 楼

我写了一个，不知道可用不，先用DP找到最长subsequence的长度，complexity is O(
nlongn);
再用recursive找到所有可行解。
public static List> AllLongestIncreasingSubsequence(int[] nums
) {
int[] m = new int[nums.length + 1];
int L = 0;
for (int i = 0; i < nums.length; i++) {
int lo = 1;
int hi = L;
while (lo <= hi) {
int mid = (lo + hi) / 2;
if (nums[m[mid]] < nums[i]) lo = mid + 1;
else hi = mid - 1;
}
int newL = lo;
m[newL] = i;
if (newL > L) {
L = newL;
}
}
List> res = new ArrayList<>();
Helper(nums, L, 0, new ArrayList(), res);
return res;
}
public static void Helper(int[] nums, int len, int index, List tmp,
List> res) {
if (tmp.size() == len) {
res.add(new ArrayList(tmp));
return;
}
for (int i = index; i < nums.length; i++) {
if (tmp.size() == 0 || tmp.get(tmp.size() - 1) < nums[i]) {
tmp.add(nums[i]);
Helper(nums, len, i + 1, tmp, res);
tmp.remove(tmp.size() - 1);
}
}
}

public static void main(String[] args) {
int[] input = {2,7,1,3,10,4,5,6};
int[] res = LongestIncreasingSubsequence(input);
for (int i : res) {
System.out.print(i + ",");
}
int[] input2 = {1,6,2,7,3,8,9,10,4,5};
List> res2 = AllLongestIncreasingSubsequence(input2);
for (List l : res2) {
for (Integer i : l) {
System.out.print(i + ",");
}
System.out.println();
}
}

B*h2011-07-29 07:07

6 楼

3折的那种915刀，2折760刀

s*x2011-07-29 07:07

7 楼

nums
Thanks a million !!

【在 k****r 的大作中提到】

: 我写了一个，不知道可用不，先用DP找到最长subsequence的长度，complexity is O(
: nlongn);
: 再用recursive找到所有可行解。
: public static List> AllLongestIncreasingSubsequence(int[] nums
: ) {
: int[] m = new int[nums.length + 1];
: int L = 0;
: for (int i = 0; i < nums.length; i++) {
: int lo = 1;
: int hi = L;

S*12011-07-29 07:07

8 楼

谢谢MM哈，，，确实好贵额。。

【在 B*********h 的大作中提到】

: 3折的那种915刀，2折760刀