美国哪里能买到安瓶？ - 未名空间MITBBS历史存档

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美国哪里能买到安瓶？

美国哪里能买到安瓶？# Fashion - 美丽时尚

G*G2010-04-12 07:04

1 楼

ems from china to u.s.
Nov 13 it arrives at new york customs.
still have not any change in the tracking information.

T*72010-04-12 07:04

2 楼

// Given a set of candidate numbers (C) and a target number (T),
// find all unique combinations in C where the candidate numbers sums to T.
//
// The same repeated number may be chosen from C unlimited number of times.
//
// Note:
// All numbers (including target) will be positive integers.
// Elements in a combination (a1, a2, … ,ak) must be in non-descending
order. (ie, a1 ≤ a2 ≤ … ≤ ak).
// The solution set must not contain duplicate combinations.
// For example, given candidate set 2,3,6,7 and target 7,
// A solution set is:
// [7]
// [2, 2, 3]

j*g2010-04-12 07:04

3 楼

国内拍婚纱照用的安瓶，哪个牌子好呢？在美国哪里能买到？

h*o2010-04-12 07:04

4 楼

包裹里有啥？一个女朋友？

k*r2010-04-12 07:04

5 楼

back track~
recursively find the right combination
void findCom(int* s, int index, vector> result, int target,
vector tmp_sum)
if (sum(tmp_sum)==target) result.push_back(tmp_sum);
else if (sum(tmp_sum)< target) {
tmp_sum.push_back(s[index]);
findCom(s, index, result, target, tmp_sum);
tmp_sum.pop_back();
findCom(s, index+1, result, target, tmp_sum);
}

G*G2010-04-12 07:04

6 楼

no. but how long would it take?

【在 h*****o 的大作中提到】

: 包裹里有啥？一个女朋友？

t*y2010-04-12 07:04

7 楼

DP
public List> CombinationSum(int[] input, int value)
{
if (value <= 0) return null;
Array.Sort(input);
List>[] tmp = new List>[value + 1];
tmp[0] = new List>();
for (int i = 1; i <= value; i++)
{
List> res = new List>();
for (int j = 0; j < input.Length; j++)
{
if (i - input[j] > 0)
{
if (tmp[i - input[j]] != null && tmp[i - input[j]].
Count != 0)
{
foreach (List l in tmp[i - input[j]])
{
// this line make sure the result is in a
increase order.
//but can select same value. [2,2,3] is
allowed.
//[2,3,2] is not allowed.
if (input[j] >= l.Last())
{
List newList = new List(l);
newList.Add(input[j]);
res.Add(newList);
}
}
}
}
else if (i == input[j])
{
res.Add(new List() { input[j] });
}
else
{
// if input[j] already larget than then there
// is no need to go further.
break;
}
tmp[i] = res;
}
}
return tmp[value];
}

r*y2010-04-12 07:04

8 楼

我的当时也是耽搁了三天了才有tracking info 出来
你应该打电话去美国的usps问一下包裹的情况

p*22010-04-12 07:04

9 楼

参考CC150硬币那题

T*72010-04-12 07:04

10 楼

Which chapter?thank u.

【在 p*****2 的大作中提到】

: 参考CC150硬币那题

l*b2010-04-12 07:04

11 楼

忽然想到有负数的话就是无限多个解啦。。。

l*b2010-04-12 07:04

12 楼

第4版在recursion 那一章

【在 T******7 的大作中提到】

: Which chapter?thank u.

T*72010-04-12 07:04

13 楼

Obviously ;)

【在 l*******b 的大作中提到】

: 忽然想到有负数的话就是无限多个解啦。。。

c*a2010-04-12 07:04

14 楼

150里面9.8
第五版

T*72010-04-12 07:04

15 楼

多谢

【在 c*****a 的大作中提到】

: 150里面9.8
: 第五版

b*i2010-04-12 07:04

16 楼

瞎写的
void search(int p, int S){
// b[p]is the array for the number of appearance of candidate c[p]
if (p==N){// N is the number of candidates
if (S==T){
for(int j=0;jfor(int k=0;kcout<cout<}
return;
}
while(S<=T){
search(p+1, S);
b[p]++;
S=S+c[p];
}
b[p]=0;
}

【在 T******7 的大作中提到】

: // Given a set of candidate numbers (C) and a target number (T),
: // find all unique combinations in C where the candidate numbers sums to T.
: //
: // The same repeated number may be chosen from C unlimited number of times.
: //
: // Note:
: // All numbers (including target) will be positive integers.
: // Elements in a combination (a1, a2, … ,ak) must be in non-descending
: order. (ie, a1 ≤ a2 ≤ … ≤ ak).
: // The solution set must not contain duplicate combinations.