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joj 1815 The Smart Bunny 动态规�?详解

luis — Mon, 20 Jul 2009 09:19:00 GMT

2^nd JOJ Cup Online VContest Problem

We all know that bunny is fond of carrots. One cloudy day, he was told that there would be a violenting flood coming soon to destroy the forests. He was scared and after not thinking too much he told himself that he had to escape. He suddenly recalled that there was a temple on the top of the hill and he might shelter there until the flood's past. But unfortunately there was no food for him on the top of the hill, so he had to take his carrots away along with himself. Then he moved to the foot of the hill and stopped. There was only one way for him to get the top of the hill, that is, a long staircase. Given the number of the steps of the staircase, he asked himself:"how many different ways of strides are there for him to get the top of the hill?". Of course, because of his height, he could only stride a limited range of steps. He was smart so much so that he got the answer quickly. Do you know how he did it?

Input Specification

The input consists of several test cases, each of which occupies a line containing M(1<=M<=40) and N(1<=N<=10), where M indicates the number of the steps of the staircase and N indicates the maximal number of steps the bunny can stride once.

Output Specification

Each test case should correspond to a line in the output. Your program should print an integer which is the answer.

Sample Input

4 2
5 4

Sample Output

5
15

题意是一只兔子要到距��MؓM�Q�单位�ؓ1�Q�的地方�Q�它每步最多走N�Q�问有多��种�Ҏ��。输入M N 输出r[m][n];

解析�Q�设为r[i][j]�Q�表�C��L��d��i且每�ơ最多走j的方法数(可以没有走出那大��ؓj的那步，只是允许走那步而已)�?br>
�Ҏ��最后那一步可能走的长度，r[i-2][j]表示最后那步距��Mؓ2.r[i-j][j]表示最后那步距��Mؓj.
建立递推关系r[i][j]=r[i-1][j]+r[i-2][j]+r[i-3][j]+r[i-j][j];
//本�h觉得�q�里特别难想�Q�想��C��也觉得无法徏立递推关系呀�Q�j不是一直不变吗�Q?br>��奇的地方在于可以��or[0][j]=r[1][j]=1;当然�q�有r[i][1]=1
-----------预处�?---
for(j=0;j {

r[0][j]=1;

r[1][j]=1;

}

for(i=0;i

r[i][1]=1;

---dp----------
for(i=2;i    for(j=2;j    {
     if(i        r[i][j]=r[i][i]; //�q�点非常重要
     else
       for(k=1;k<=j;k++)
        {
   r[i][j]+=r[i-k][j];
        }
    }
------------
r[1][1]=1
r[2][1]=1
r[2][2]=r[2][1]+r[2][2]=2;
r[3][1]=1;
r[3][2]=r[2][2]+r[1][2]=3;
r[3][3]=r[0][3]+r[1][3]+r[2][3]=4;

luis 2009-07-20 17:19 发表评论

joj 1762 Dollars 动态规�?详解

luis — Sun, 19 Jul 2009 13:45:00 GMT

New Zealand currency consists of $100, $50, $20, $10, and $5 notes and $2, $1, 50c, 20c, 10c and 5c coins. Write a program that will determine, for any given amount, in how many ways that amount may be made up. Changing the order of listing does not increase the count. Thus 20c may be made up in 4 ways: 1 20c, 2 10c, 10c+2 5c, and 4 5c.

Input

Input will consist of a series of real numbers no greater than $50.00 each on a separate line. Each amount will be valid, that is will be a multiple of 5c. The file will be terminated by a line containing zero (0.00).

Output

Output will consist of a line for each of the amounts in the input, each line consisting of the amount of money (with two decimal places and right justified in a field of width 5), followed by the number of ways in which that amount may be made up, right justified in a field of width 12.

Sample input

0.20
2.00
0.00

Sample output

0.20           4
2.00         293

典型的动态规划问题：

可以先考虑只有n-1�U�硬币，V1,V2,V3,---Vn-1.讑և�成��d��gؓM的方法数为result[M];

现在又添加一�U�面��gؓVn的硬币�?/pre>
则咱们需要修改一些��|��修改那些大于�{�于Vn的result[],不妨设M大于vn�?/pre>
新result[M]=原result[M]+result[M-Vn]+result[M-2*Vn]+----直到M-p*Vn<=0;�Q?�Q?/pre>
实际上如果如(2)那样处理 从大于等于Vn的result[]开始处理的话，从小到大�Q�设Vn=M-p*Vn;
则result[M-p*Vn]�W�一个最先更斎ͼ�result[M-p*Vn]+=result[M-p*Vn -Vn]   �?span style="COLOR: red">(2)
随后更新result[M-(p-1)*vn]�Q�则只需直接加上result[M-p*Vn]的��|��无需如（1�Q�式那样  累加�?/pre>
用程序表辑ְ��?/pre>
int coin[10]={5,10,20,50,100,200,500,1000,2000,5000};
for(i=1;i<10;i++)
 for(j=coin[i];j<=50000;j+=5)
     result[j]+=result[j-Coin[i]];   (2) //
-------------
�q�有一个注意点�Q�见牛�h博客�Q?a >http://hi.baidu.com/piaoshi111/blog/item/9a6de84a00a6e5f882025c89.html
���是���点数的四舍五入�Q?/pre>
1.5�Q���Q�Ҏ��可能表�ؓ1.4999999�Q�所以乘�?00时�{化�ؓ整型�?49�Q�应当注意�?/pre>


luis 2009-07-19 21:45 发表评论

joj 1832 Little Shop of Flowers

luis — Tue, 14 Jul 2009 02:05:00 GMT

You want to arrange the window of your flower shop in a most pleasant way. You have F bunches of flowers, each being of a different kind, and at least as many vases ordered in a row. The vases are glued onto the shelf and are numbered consecutively 1 through V, where V is the number of vases, from left to right so that the vase 1 is the leftmost, and the vase V is the rightmost vase. The bunches are moveable and are uniquely identified by integers between 1 and F. These id-numbers have a significance: They determine the required order of appearance of the flower bunches in the row of vases so that the bunch i must be in a vase to the left of the vase containing bunch j whenever i < j. Suppose, for example, you have bunch of azaleas (id-number=1), a bunch of begonias (id-number=2) and a bunch of carnations (id-number=3). Now, all the bunches must be put into the vases keeping their id-numbers in order. The bunch of azaleas must be in a vase to the left of begonias, and the bunch of begonias must be in a vase to the left of carnations. If there are more vases than bunches of flowers then the excess will be left empty. A vase can hold only one bunch of flowers.

Each vase has a distinct characteristic (just like flowers do). Hence, putting a bunch of flowers in a vase results in a certain aesthetic value, expressed by an integer. The aesthetic values are presented in a table as shown below. Leaving a vase empty has an aesthetic value of 0.

		V A S E S
		1	2	3	4	5
Bunches	1 (azaleas)	7	23	-5	-24	16
	2 (begonias)	5	21	-4	10	23
	3 (carnations)	-21	5	-4	-20	20

According to the table, azaleas, for example, would look great in vase 2, but they would look awful in vase 4.

To achieve the most pleasant effect you have to maximize the sum of aesthetic values for the arrangement while keeping the required ordering of the flowers.

ASSUMPTIONS

1 ≤ F ≤ 100 where F is the number of the bunches of flowers. The bunches are numbered 1 through F. F ≤ V ≤ 100 where V is the number of vases. -50 ≤ Aij ≤ 50 where Aij is the aesthetic value obtained by putting the flower bunch i into the vase j.

Input

The first line contains two numbers: F and V.

The following F lines: Each of these lines contains V integers, so that Aij is given as the j’th number on the (i+1)’st line of the input file.

Notice: The input contains several test cases.

Output

The output line will contain the sum of aesthetic values for your arrangement.

Sample Input

3 5
7 23 -5 -24 16
5 21 -4 10 23
-21 5 -4 -20 20

Sample Output

�q�题可以用搜索过�Q�但是还可以用dp
用result[i][j]表示前i行，以j�l�尾的排法的最大��|��
rsult[1][j]直接初始化�ؓnum[i][j];其余初始化�ؓ负无�I?br>dp的过�E�就�?br>    for(i=2;i<-r;i++)//行逐渐增加
          for(j=i;j<=c;j++)//列必��d��于等于行��P��否则无法保证从左上方到右下方
                    for(k=1;k                               if(result[i][j]                                               result[i][j]=result[i-1][k]+num[i][j]
更详�l�的代码可以到苏强的博客http://download.csdn.net/user/china8848/获得

luis 2009-07-14 10:05 发表评论

poj 1631 Bridging signals �l�典dp 旉��复杂度nlogn

luis — Sun, 12 Jul 2009 07:46:00 GMT

题意��是求最长上升子序列�Q�且必须用nlogn的dp��法�?br>

技巧：讄��一个数�l�a[i]

存放所有长度�ؓi的上升子序列中最��的末元素��|��比如说只有两个长度�ؓ3的上升子序列123�?24�Q�那么a[3]中存攄��是3�Q�末元素3<4�Q?/p>

那么当来一个新数data�Ӟ��如果它的值大于最镉K��度的末元素的��|��即a[ans]�Q�，则ans++�Q�且a[ans]=data;

否则�Q�通过二分查找�Q�数�l�a中的元素为递增�Q�，��最接近data且大于data的那个元素更��Cؓdata�Q�既最��的大于它的数�?br>例如1,5,3,4,之后来个2�Q�a[1]=1,a[2]=3,a[3]=4;则更新a[2]=2;
�׃��二分查找复杂度�ؓlog(n)�Q�外围�ؓn,�ȝ��复杂度�ؓnlogn

代码�?

#include <stdio.h>
int res[40000];
int binSearch(int left, int right, int num)//扑ֈ�最��的大于�{�于它的�?nbsp;
{
    while(left <= right)
    {
        int mid=(left + right)/2;
         if(res[mid]<num)
            left=mid+1;
        else
            right=mid-1;
    }
    return right;
}

int main()
{
//freopen("s.txt","r",stdin);
// freopen("key.txt","w",stdout);
    int t, n, num;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d %d", &n, &num);
        res[0]=num;
        int tot = 1;
        for(int i=1;i<n;++i)
        {
            scanf("%d", &num);
            if(num>=res[tot-1])
            res[tot++]=num;
            else
            {
                int pos=binSearch(0,tot-1,num);//扑ֈ�最��的大于它的�?nbsp;
                res[pos+1] = num;
            }
        }
        printf("%d\n", tot);
    }
    return 0;
}

luis 2009-07-12 15:46 发表评论

joj 1092 To the Max 转帖Dp�l�典

luis — Thu, 09 Jul 2009 10:37:00 GMT

[原创]POJ1050 To the Max 解题报告

2006-04-17 11:55

题目大意:

��d��一个n*n的数�l?比如

0 -2 -7 0

9 2 -6 2

-4 1 -4 1

-1 8 0 -2

从里面�Q意截取一个矩�?使得矩阵所包含的数字的和最�?

截取出来的矩�?和�ؓ15

9 2

-4 1

-1 8

---------------------------------------------------------

POJ 1050 我的解题报告�Q?br>
�q�个题目很经典的��_��O�Q�N^3�Q�的DP�?br>
首先偶们考察�q�样的题目，��化版�Q?br>
已知一列数�Q�求��L��q�箋若干个数和的最大倹{�?br>
SAMPLE�Q?nbsp;3 2 -6 2 -1 7

原数3        2      -6       2      -1       7

处理3        5      -1       2       1       8

因�ؓ是连�l�若�q�个自然数的和，那么�Q�前面的某个数字取与不取的条件在于：以前面这个数字�ؓ�l�尾的连�l�数的和最大值是否大�?�Q�如果大�?�Q�那么这个数字必然要会出现在包括数字的序列中�Q�否则无法做到最大�?br>
所以，昄��。处理的原则是maxn[i]=max{0,maxn[i-1]}+a[i];

�׃��无须记录位置。所以，可以直接用一个变量sum代替maxn数组。O(n)的扫描即可�?br>
单列数字的问题解决了�Q�下面我们考察多列数字�?br>
sample:

         0    -2    -7    0

         9     2    -6    2

        -4     1    -4    1

        -1     8     0   -2

我们可以��多列数字�{换成单列数字来做�Q?nbsp;可以�q�样设想�Q�结果是一个长方�Ş�Q�我们把他压扁，使得宽�ؓ1�?br>
引入辅助数组st,st[i][j]代表�W�i列从�W?行开始的数字累加到第j行的倹{��那么，我们每次压扁的时候，��可以用st[i][j]-st[i][k-1]来表�C�第i列从�W�k个数字篏加到�W�j个数字的倹{��达到压�~�的效果。然后用上面单列数字的方法来做。算法时间复杂度O (N^3)

Source

Problem Id:1050  User Id:galaxy

Memory:112K  Time:0MS

Language:G++  Result:Accepted

/*

  Name:POJ 1050

  Copyright: flymouse@galaxy

  Author:chenlei

  Date: 15-02-06 07:36

  Description: DP O(N^3)

*/

#include

#include

#define mt 101

int main()

{

int a[mt][mt];

int st[mt][mt];

int p,k,n,i,j,sum,maxn;

//freopen("in.txt","r",stdin);

scanf("%d",&n);

for (i=1;i<=n;i++)

for (j=1;j<=n;j++)

scanf("%d",&a[i][j]);

memset(st,0,sizeof(st));

for (i=1;i<=n;i++)

   for (j=1;j<=n;j++)

  st[i][j]=st[i][j-1]+a[j][i];

  maxn=0;

   for (i=1;i<=n;i++)

   {

for (j=i;j<=n;j++)

{

p=st[1][j]-st[1][i-1];

sum=p;

for (k=2;k<=n;k++)

{

if (sum>0)

sum+=st[k][j]-st[k][i-1];

else sum=st[k][j]-st[k][i-1];

if (sum>p) p=sum;

}

if (p>maxn) maxn=p;

}

   }

   printf("%d\n",maxn);

   return 0;

原文地址�Q?a >http://hi.baidu.com/flymouse/blog/item/fd1378f05c7ff7c37931aac3.html

luis 2009-07-09 18:37 发表评论

luis — Sat, 04 Jul 2009 01:27:00 GMT

Energy

Status	In/Out	TIME Limit	MEMORY Limit	Submit Times	Solved Users	JUDGE TYPE
	stdin/stdout	3s	10240K	717	196	Standard

Mr. Jojer is a very famous chemist. He is doing a research about behavior of a group of atoms. Atoms may have different energy and energy can be positive or negative or zero, e.g. 18 or -9. Absolute value of energy can not be more than 100. Any number of continuous atoms can form an atom-group. Energy of an atom-group is defined by the sum of energy of all the atoms in the group. All the atoms form an atom-community which is a line formed by all the atoms one by one. Energy of an atom-community is defined by the greatest energy of an atom-group that can be formed by atoms in the atom-community. The problem is, given an atom-community, to calculate its energy.

Input

The input contains several test cases. Each test case consists of two lines describing an atom-community. The first line of each test case contains an integer N(N<=1000000), the number of atoms in the atom-community. The second line of each test case contains N integers, separated by spaces, each representing energy of an atom, given in the order according to the atom-community. The last test case marks by N=-1, which you should not proceed.

Output

For each test case(atom-community description), print a single line containing the energy.

Sample Input

5
8 0 6 4 -1
-1

Sample Output

理解题意很重要，题目的意思是�?在m个中�?nbsp; �q�箋的n个atom能量�?使其最大�?br>�E�序中sumtemp�Q�sum。sumtemp��定的是其左边界�Q�sum��定其右边界�?br>sumtemp��定左边前n个数之和��的最大的n�Q�且�W�n个数昄��Q�然后从n+1开始选数�?br>sum��定了右边界使其最大�?br>求和最大都可以用这�U�思�\�Q�！�Q�！�Q�！�Q�！�Q�！
举例
1�Q?�Q?4�Q?�Q?�Q?2�Q?�Q?�Q?6�Q?�Q?3�Q?�Q?0
��L��sumt1=1,sum=1
sumt2=2;sum=1+2=3;
sum3=sum2-4=-1;则sumtemp=0;sum不变�?br>sum4=sumtemp+4;sumsum5=sumtemp+2�Q�sum��M��sum只有在sum一�l�dp.
#include"stdio.h"
int main()
{
freopen("s.txt","r",stdin);
freopen("key.txt","w",stdout);
int n;
int a;
while(scanf("%ld",&n),n!=-1)
{
      long long sum=-0x7fffffff,sumtemp=-0x7fffffff;
   for(long i=0;i   {
    scanf("%d",&a);
    if(sumtemp>0)
   sumtemp+=a;
    else
   sumtemp=a;
    if(sumtemp>sum)
   sum=sumtemp;

}
printf("%lld\n",sum);

}
return 0;
}

luis 2009-07-04 09:27 发表评论

luis — Thu, 02 Jul 2009 08:05:00 GMT

Tug of War

Status	In/Out	TIME Limit	MEMORY Limit	Submit Times	Solved Users	JUDGE TYPE
	stdin/stdout	15s	8192K	479	114	Standard

A tug of war is to be arranged at the local office picnic. For the tug of war, the picnickers must be divided into two teams. Each person must be on one team or the other; the number of people on the two teams must not differ by more than 1; the total weight of the people on each team should be as nearly equal as possible.

The first line of input contains n the number of people at the picnic. n lines follow. The first line gives the weight of person 1; the second the weight of person 2; and so on. Each weight is an integer between 1 and 450. There are at most 100 people at the picnic.

The input may contain several test cases.

Your output will be a single line containing 2 numbers: the total weight of the people on one team, and the total weight of the people on the other team. If these numbers differ, give the lesser first.

Sample Input

Output for Sample Input

190 200

利用dp思想 �Q�n为偶数时求出s(n,n/2)�Q�n为奇数时也是s(2n,n/2)�Q�和sum/2最接近的那个。非常经典的思�\�?br>S(k, 1) = {A[i] | 1<= i <= k}
S(k, k) = {A[1]+A[2]+…+A[k]}
S(k, i) = S(k-1, i) U {A[k] + x | x属于S(k-1, i-1) }
//一下代码只能用于sum特别��的情况�Q�否则会��时�Q�！�Q�！�Q�！�Q�！�Q�！�Q?br>#include
#include
#define MAX 101
#define min(a,b) ((a)<(b) ? (a) : (b))
using namespace std;

int main()
{
freopen("s.txt","r",stdin);
freopen("key.txt","w",stdout);
int num;
int a[MAX],i,j,k,l,m,NUM;
bool s[MAX][2500];
while(cin>>num)
{
  int sum=0;
    for(i=1;i<=num;i++)
   {
cin>>a[i];
sum+=a[i];
   }
   if(num%2==0)NUM=num/2;
   else NUM=num/2+1;

    for(i=0;i<=num;i++)
     for(j=0;j<=sum/2;j++)
      s[i][j]=false;//表示取i个物品能否达到重量是j.

   s[0][0]=true;
   for(k=1;k<=num;k++)//必须在最外层�Q�元素不能重�?
   for(j=min(k,NUM);j>=1; j--)//递减的结果是使得不会出现在同一层次的互为因�?、、、、、、、、、、、��y妙的实现了课本上的序偶对生成法�?br>   for(i=a[k];i<=sum/2;i++)
   {
  if(s[j-1][i-a[k]])
  s[j][i]=true;
}

for(i=sum/2; i>=0; i--) {
    if(s[NUM][i]) {
        cout <        break;
    }
}

}
//system("PAUSE");
return 0;
}
下一�ơ实��C��个序偶生成法�?/p>

#include
#include
using namespace std;

int a[101];
bool b[101][45002];

int main(){
// freopen("s.txt","r",stdin);
// freopen("key.txt","w",stdout);
    int N,M,i,j,k;
    while(scanf("%d",&N)!=EOF){
         memset(b,0,sizeof(b));
         a[0]=M=0;
        for(i=1;i<=N;i++){
             scanf("%d",a+i);
             M+=a[i];
         }
         b[0][0]=1;
        for(k=1;k<=N;k++){
            for(i=1;i<=N/2;i++){
                for(j=M/2;j>=0;j--){
                    if(b[i-1][j]){
                         b[i][j+a[k]]=1;
                     }
                 }
             }
         }
        for(i=M/2,j=M/2+1;i>=0;i--,j++){
            if(b[N/2][i]){
                 printf("%d %d\n",i,M-i);
                break;
             }
            if(b[N/2][j]){
                 printf("%d %d\n",M-j,j);
                break;
             }
         }
     }
    return 0;
}

luis 2009-07-02 16:05 发表评论

joj 2521 Monkey and fruits 矛_��合�ƈ问题

luis — Tue, 30 Jun 2009 13:01:00 GMT

After culling his most favorite fruits from the trees in an orchard,Keen AS, a mischievous monkey, is confused of the way in which he can combine these piles of fruits into one in the most laborsaving manner. Each turn he can merely combine any two piles into a larger one. Obviously, by combining N-1 times for N piles of fruits, one pile results eventually. It always takes AS a certain quantity of units of stamina, which equals to the sum of the amounts of fruits in each of the two piles combined, to complete a combination. For instance, given three piles of fruits, containing 1, 2, and 9 fruits respectively, the combination of 1 and 2 would cost 3=1+2 units of stamina, and two piles-3 and 9-are resulted; finally, combining them would cost another 12=3+9 units of stamina, and the total units of stamina taken in the whole procedure is 15=3+12. Surely there are many a possible means to combine these three piles; however, it can be proved that 15 is the minimum amount of units of stamina in demand, and that is what your program is required to do. in this problem ,the fruits are put in a strait line. A pile of fruits 'P' can only combined with the rightest pail on the left of P or the leftest on the right of P, and the united pile will at the position of the bigger one.

Input

The input file consists of many test cases. The first line contains an integer N (<=100), indicating the number of piles, and the second line contains N integers, each of which represents the amount of fruits in a pile. no integer will more than 10000, the sequence of integer also means the location of the piles.

Output

Your program should print the minimum amount of units of stamina that are required to combine these piles of fruits into one on request.

Sample Input

3
1 2 9

Sample Output

典型的石子合�q��题�?br>#include
int a[101][101],sum[101];
int stone[101];
int main()
{
    int n,i,j,k,flag,l,t;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=1;i<=n;i++)
        scanf("%d",&stone[i]);
        sum[0]=0;
        for(i=1;i<=n;i++)
        {
            sum[i]=sum[i-1]+stone[i];
            a[i][i]=0;
        }
        for(l=1;l<=n-1;l++)
        for(i=1;i<=n-l;i++)
        {
            flag=0;
            j=i+l;
            for(k=i;k            {
                t=a[i][k]+a[k+1][j]+sum[j]-sum[i-1];
                if(!flag)
                {
                    flag=1;
                    a[i][j]=t;
                }
                else if(t                a[i][j]=t;
            }
        }
        printf("%d\n",a[1][n]);
    }
    return 0;
}

luis 2009-06-30 21:01 发表评论

luis — Sat, 27 Jun 2009 02:44:00 GMT

One curious child has a set of N little bricks. From these bricks he builds different staircases. Staircase consists of steps of different sizes in a strictly descending order. It is not allowed for staircase to have steps equal sizes. Every staircase consists of at least two steps and each step contains at least one brick. Picture gives examples of staircase for N=11 and N=5:

Your task is to write a program that reads from input numbers N and writes to output numbers Q - amount of different staircases that can be built from exactly N bricks.

Input

Numbers N, one on each line. You can assume N is between 3 and 500, both inclusive. A number 0 indicates the end of input.

Output

Numbers Q, one on each line.

Sample Input

3
5
0

Sample Output

1
2

�Ҏ��1�Q�动态规�?br>
#include
#include
using namespace std;
int main()
{
freopen("s.txt","r",stdin);
freopen("key.txt","w",stdout);
double f[501][501]={0};
double s;
int i,j,k,n;
for(i=3;i<=500;i++)
for(j=1;j<=(i-1)/2;j++)
    f[i][j]=1;
for(i=3;i<=500;i++)
    for(j=1;j<=(i-1)/2;j++)
    {
for(k=j+1;k<=(i-j-1)/2;k++)
   f[i][j]=f[i-j][k];
}
    while(scanf("%d",&n),n) {
        s=0;
        for(i=1;i<=(n-1)/2;i++) s+=f[n][i]; //f[n]=f[n][1]+f[n][2]+-----+f[n][floor((i-1)/2)]
        printf("%.0f\n",s);
    }
//system("PAUSE");
return   0;
}
更妙的方法：生成函数�?br>计算(1+x)(1+x^2)(1+x^3)-----,x^n的系数即为所�?br>int i,j;
double ans[510]={1,1};//已经把ans[1]和ans[0]赋�ؓ1了，其余�?
for(i=2;i<=500;i++) {
        for(j=500;j>=0;j--) {
            if(i+j<=500) ans[i+j]+=ans[j];
        }
    }

先计��?1+x)(1+x^2)
再计��?1+x)(1+x^2)   *(1+x^3)

luis 2009-06-27 10:44 发表评论

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