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pku1309 数学优化+枚�D

yzhw — Mon, 18 Jul 2011 16:10:00 GMT

题目

Coconuts, Revisited

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 1832		Accepted: 737

Description

The short story titled Coconuts, by Ben Ames Williams, appeared in the Saturday Evening Post on October 9, 1926. The story tells about five men and a monkey who were shipwrecked on an island. They spent the first night gathering coconuts. During the night, one man woke up and decided to take his share of the coconuts. He divided them into five piles. One coconut was left over so he gave it to the monkey, then hid his share and went back to sheep.

Soon a second man woke up and did the same thing. After dividing the coconuts into five piles, one coconut was left over which he gave to the monkey. He then hid his share and went back to bed. The third, fourth, and fifth man followed exactly the same procedure. The next morning, after they all woke up, they divided the remaining coconuts into five equal shares. This time no coconuts were left over.

An obvious question is "how many coconuts did they originally gather?" There are an infinite number of answers, but the lowest of these is 3,121. But that's not our problem here.

Suppose we turn the problem around. If we know the number of coconuts that were gathered, what is the maximum number of persons (and one monkey) that could have been shipwrecked if the same procedure could occur?

Input

The input will consist of a sequence of integers, each representing the number of coconuts gathered by a group of persons (and a monkey) that were shipwrecked. The sequence will be followed by a negative number.

Output

For each number of coconuts, determine the largest number of persons who could have participated in the procedure described above. Display the results similar to the manner shown below, in the Expected Output. There may be no solution for some of the input cases; if so, state that observation.

Sample Input

25 30 3121 -1

Sample Output

25 coconuts, 3 people and 1 monkey 30 coconuts, no solution 3121 coconuts, 5 people and 1 monkey

Source

North Central North America 1997

解法�Q?br />首先写出递推公式
f(0)=A A=nk
f(i)=f(i-1)/(n-1)*n+1

随便什么方法写出闭形式
f(n)=[�Q�n^n�Q?�Q�A+n-1�Q�]/[�Q�n-1�Q�^n]-(n-1)
题目中告诉f(n)的��|��求n最大�?br />首先观察下前面那个分式，�׃��n和n-1互质�Q�所以n^n�?n-1)^n也互�?/span>�Q�分式结果要��Z��个整敎ͼ�f(n)+n-1中必��d��有因子n^n�Q�换句话��_��f(n)+n-1>n^n,题目中给的f(n)可以�?2位整数表�C�，那么n必然��于12�Q?/span>
下面不用说什么了�Q�暴力吧�Q�肯�?MS了~不过��Z��完美�Q�n^n我用了二�q�制快速幂~具体看代码吧

代码�Q?br />

1 Source Code
2 Problem: 1309        User: yzhw
3 Memory: 392K        Time: 0MS
4 Language: G++        Result: Accepted
5
6     Source Code
7
8     # include <cstdio>
9     using namespace std;
10     long long pow(int a,int b)
11     {
12         long long ans=1,t=a;
13         while(b)
14         {
15             if(b&1) ans*=t;
16             t*=t;
17             b>>=1;
18         }
19         return ans;
20     }
21     int main()
22     {
23         //freopen("input.txt","r",stdin);
24         int n;
25         while(scanf("%d",&n)!=EOF&&n>=0)
26         {
27             int ans=-1,i;
28             for(i=2;i<=12;i++)
29             {
30                 long long t=n;
31                 t+=i-1;
32                 long long t1=pow(i,i),t2=pow(i-1,i);
33                 if(t%t1==0)
34                 {
35                     t=t/t1*t2-i+1;
36                     if(t>=0&&t%i==0) ans=i;
37                 }
38             }
39             if(ans==-1) printf("%d coconuts, no solution\n",n);
40             else printf("%d coconuts, %d people and 1 monkey\n",n,ans);
41         }
42         return 0;
43     }
44
45

yzhw 2011-07-19 00:10 发表评论

pku1894-1903 Northeastern Europe 2003 解题报告

yzhw — Thu, 03 Feb 2011 17:08:00 GMT

摘要: Solved ID Title Ratio(AC/att) Yes Problem A Alternative Scale of Notation ... 阅读全文

yzhw 2011-02-04 01:08 发表评论

pku2894-2903 Tehran 2005 比赛�ȝ��

yzhw — Sun, 30 Jan 2011 17:59:00 GMT

摘要: Solved ID Title Ratio(AC/att) Yes Problem A Ancient Keyboard 10... 阅读全文

yzhw 2011-01-31 01:59 发表评论

yzhw — Thu, 02 Dec 2010 16:13:00 GMT

题意是这�?br>N个集装箱�Q�俯视图如下�Q�最多叠5层。要求底面积最��。如果有重复�Ҏ��Q�要求长、宽的绝对值差最��?br>
��寸�Q�集装箱 8*40 ,水��^间距�Q?�Q�竖直间距：2

题解�Q?br>枚�D�U�向个数�Q�枚丑ֈ�sqrt(上取��_��n/5)卛_��?br>
�E�序�Q?br>

1import java.util.*;
2public class Main {
3
4    /**
5     * @param args
6     */
7    public static void main(String[] args) {
8        Scanner in=new Scanner(System.in);
9        int test=in.nextInt();
10        while((test--)!=0)
11        {
12            long num=in.nextLong();
13            num=(num%5==0?num/5:num/5+1);
14            long r1=Long.MAX_VALUE,r2=Long.MAX_VALUE,l=0,w = 0;
15            for(long a=1;a*a<=num;a++)
16            {
17                long b=(num%a==0?num/a:num/a+1);
18                if((a*44+4)*(b*10+2)<r1||(a*44+4)*(b*10+2)==r1&&Math.abs((a*44+4)-(b*10+2))<r2)
19                {
20                    r1=(a*44+4)*(b*10+2);
21                    r2=Math.abs((a*44+4)-(b*10+2));
22                    l=(a*44+4);
23                    w=(b*10+2);
24                }
25                if(b==1) break;
26            }
27            System.out.println(Math.max(w,l)+" X "+Math.min(w, l)+" = "+r1);
28        }
29
30    }
31
32}
33

yzhw 2010-12-03 00:13 发表评论

yzhw — Mon, 15 Nov 2010 16:50:00 GMT

Error Curves

Time Limit: 2 Seconds Memory Limit: 65536 KB

Josephina is a clever girl and addicted to Machine Learning recently. She pays much attention to a method called Linear Discriminant Analysis, which has many interesting properties.

In order to test the algorithm's efficiency, she collects many datasets. What's more, each data is divided into two parts: training data and test data. She gets the parameters of the model on training data and test the model on test data.

To her surprise, she finds each dataset's test error curve is just a parabolic curve. A parabolic curve corresponds to a quadratic function. In mathematics, a quadratic function is a polynomial function of the form f(x) = ax² + bx + c. The quadratic will degrade to linear function if a = 0.

It's very easy to calculate the minimal error if there is only one test error curve. However, there are several datasets, which means Josephina will obtain many parabolic curves. Josephina wants to get the tuned parameters that make the best performance on all datasets. So she should take all error curves into account, i.e., she has to deal with many quadric functions and make a new error definition to represent the total error. Now, she focuses on the following new function's minimal which related to multiple quadric functions.

The new function F(x) is defined as follow:

F(x) = max(S_i(x)), i = 1...n. The domain of x is [0, 1000]. S_i(x) is a quadric function.

Josephina wonders the minimum of F(x). Unfortunately, it's too hard for her to solve this problem. As a super programmer, can you help her?

Input

The input contains multiple test cases. The first line is the number of cases T (T < 100). Each case begins with a number n(n ≤ 10000). Following n lines, each line contains three integers a (0 ≤ a ≤ 100), b (|b| ≤ 5000), c (|c| ≤ 5000), which mean the corresponding coefficients of a quadratic function.

Output

For each test case, output the answer in a line. Round to 4 digits after the decimal point.

Sample Input

Sample Output

0.0000
0.5000

���明题意：求一�?span style="COLOR: red">开口向�?/span>的二�ơ函数在[0,1000]范围上函数值最大值的最���倹{�?/pre>
二次函数的子集仍然�ؓ凸函敎ͼ�所以可以用三分法求极倹{��精度实在很蛋疼�Q�这题要求值域�_�����?e-4�Q�但是定义域没说�_����到多���，�l�果死wa�Q�卡�?e-10�l�于�q�了。�?/pre>
贴代�?/pre>
 1# include <cstdio>
 2# include <cmath>
 3using namespace std;
 4int n;
 5int data[10001][3];
 6# define max(a,b) ((a)>(b)?(a):(b))
 7double cal(double mid)
 8{
 9   double res=-1e26;
10   for(int i=0;i<n;i++)
11     res=max(res,data[i][0]*mid*mid+data[i][1]*mid+data[i][2]);
12   return res;
13}
14int main()
15{
16    int test;
17    scanf("%d",&test);
18    while(test--)
19    {
20       scanf("%d",&n);
21       for(int i=0;i<n;i++)
22         scanf("%d%d%d",&data[i][0],&data[i][1],&data[i][2]);
23       double s=0.0,e=1000.0;
24       double last=s;
25       while(fabs(e-s)>1e-10)
26       {
27       
28         double m1=(s+e)/2.0,m2=(m1+e)/2.0;
29         if(cal(m1)<cal(m2))
30           e=m2;
31         else 
32           s=m1;
33       }
34       printf("%.4lf\n",cal(e));
35    }
36    return 0;
37}
38
39



yzhw 2010-11-16 00:50 发表评论

yzhw — Tue, 19 Oct 2010 16:36:00 GMT

题目说的有点诡异�Q�一开始没惛_��二分�Q�想直接通过半��^面交来解或者是�U�性规划。。�?br>描述下题目��^面上有一些点�Q�x,t�Q�（题目中描�q�的�?t,x)�Q�然后t是纵坐标�Q�有点不舒服�Q�，然后满��t2 >= t1+|x2-x1|的点�Q�t1,x1�Q�称为点(x2,t2)的前导点。现在知道��^面点集以及其前导点的个数�Q�问其前导点集中t的最大倹{�?br>�Ҏ��不等式，可以发现每个点的前导炚w��范围是一个以该点为顶点的三角形区域（�Ҏ��不等式约束画个图看看��明白了�Q�，然后可以通过二分求得t�Q�关于判断当前t的可行性，��每个点表示为合法区间[x2-(t2-t1),x2+t2-t1]�Q�然后求�q�些区间的最��覆盖数�Q�就是求得一些子区间�Q��得这些子区间包含于原区间�Q�。这个问题可以通过贪心��法�Q�以区间右端点�ؓ�W�一关键字，左端点�ؓ�W�二关键字进行排序，然后每次贪心的选择当前区间的右端点作�ؓ子区��_��l�计要导出全部区间需要子区间的个数num�?br>代码如下�Q?br>

1 import java.io.*;
2 import java.util.Arrays;
3 public class Main {
4
5     /**
6      * @param args
7      */
8     static StreamTokenizer in=new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
9     //static StreamTokenizer in=new StreamTokenizer(new BufferedReader(new InputStreamReader(new FileInputStream("input.txt"))));
10     static final int readInt() throws Exception
11     {
12         in.nextToken();
13         return (int)in.nval;
14     }
15     static class node implements Comparable<node>
16     {
17         int t,x,s,e;
18         public int compareTo(node pos)
19         {
20             if(e!=pos.e) return e-pos.e;
21             else return s-pos.s;
22         }
23     }
24     static node data[]=new node[100005];
25     public static void main(String[] args) throws Exception{
26         int n=readInt();
27         for(int i=0;i<100005;i++)
28             data[i]=new node();
29         for(int test=1;test<=n;test++)
30         {
31             int num=readInt(),limit=readInt(),e=4000000,s=-4000000;
32             for(int i=0;i<num;i++)
33             {
34                 data[i].t=readInt();
35                 data[i].x=readInt();
36                 e=Math.min(e, data[i].t);
37             }
38             while(s<=e)
39             {
40                 int mid=(s+e)>>1,count=1;
41                 for(int i=0;i<num;i++)
42                 {
43                     data[i].s=data[i].x-data[i].t+mid;
44                     data[i].e=data[i].x+data[i].t-mid;
45                 }
46
47                 Arrays.sort(data,0,num);
48                 int laste=data[0].e;
49                 for(int i=1;i<num;i++)
50                     if(data[i].s<=laste)
51                         continue;
52                     else
53                     {
54                         count++;
55                         laste=data[i].e;
56                     }
57                 if(count<=limit)
58                     s=mid+1;
59                 else
60                     e=mid-1;
61             }
62             System.out.println("Case "+test+": "+e);
63         }
64
65     }
66
67 }
68

yzhw 2010-10-20 00:36 发表评论

pku 1176 Party Lamps

yzhw — Tue, 19 Oct 2010 06:17:00 GMT

摘要: �q�道题题意是�Q�有N盏灯�Q�每个灯有两个状态：开、关。有4个按钮，�W�一个按钮��得所有灯改变状态，�W�二个按钮��得奇数号灯改变状态，�W�三个按钮��得偶数号灯改变状态，�W�四个按钮��?K+1��L��改变状态。开始所有灯都是亮着的，�l�出操作�ơ数�Q�最后亮着的灯和灭了的灯，求最后所有可能的状态。这题可以用模二方程�l�来表示。设a、b、c、d分别为第一个、第二个、第三个、第四个按钮按过的次数。count为操作��L��Q�满��I��?.. 阅读全文

yzhw 2010-10-19 14:17 发表评论

yzhw — Tue, 12 Oct 2010 15:01:00 GMT

题目很长�Q�就是说用从n�U�硬币中选取4�U�不同的��币堆叠��h��Q�作为桌子的四个腿，昄��Q�桌子的四个腿要相等。问对于一个给定的高度h�Q�硬币能够堆成的不大于h的最大高度和不小于h的最��高度分别�ؓ多少�Q?br>应�ؓ数据不大�Q?n�?0左右�Q�可以用4个for循环来枚丄��币，假设四种��币为a、b、c、d�Q�求的这四个数的最��公倍数卛_��Q�然后就用欧几里得算法解冟�?br>

1/*
2Source Code
3
4Problem: 1855  User: yzhw
5Memory: 8648K  Time: 313MS
6Language: G++  Result: Accepted
7*/
8
9
10# include <iostream>
11# include <cstdio>
12using namespace std;
13int l[51];
14int refer[51][51][51][51];
15int gcd(int a,int b)
16{
17    while(b)
18    {
19        int t=a%b;
20        a=b;
21        b=t;
22    }
23    return a;
24}
25int main()
26{
27    int n,t;
28    while(true)
29    {
30        scanf("%d%d",&n,&t);
31        if(!n&&!t) break;
32        for(int i=0;i<n;i++)
33            scanf("%d",l+i);
34        for(int a=0;a<n;a++)
35                                for(int b=a+1;b<n;b++)
36                                    //if(a!=b)
37                                    for(int c=b+1;c<n;c++)
38                                        //if(a!=c&&b!=c)
39                                        for(int d=c+1;d<n;d++)
40                                            //if(a!=d&&b!=d&&c!=d)
41                                            {
42                                                int lcd1=l[a]*l[b]/gcd(l[a],l[b]),lcd2=l[c]*l[d]/gcd(l[c],l[d]);
43                                                int lcd=lcd1*lcd2/gcd(lcd1,lcd2);
44                                                refer[a][b][c][d]=lcd;
45
46                                            }
47
48        while(t--)
49        {
50            int minnum=0xfffffff,maxnum=-1;
51            int pos;
52            scanf("%d",&pos);
53            for(int a=0;a<n;a++)
54                for(int b=a+1;b<n;b++)
55                    for(int c=b+1;c<n;c++)
56                        for(int d=c+1;d<n;d++)
57                            {
58
59                                int lcd=refer[a][b][c][d];
60                                if(pos-pos%lcd>maxnum)
61                                    maxnum=pos-pos%lcd;
62                                if(pos%lcd&&lcd*(pos/lcd+1)<minnum)
63                                    minnum=lcd*(pos/lcd+1);
64                                if(pos%lcd==0&&pos<minnum)
65                                    minnum=pos;
66
67                            }
68            printf("%d %d\n",maxnum,minnum);
69        }
70    }
71    return 0;
72}
73
74

yzhw 2010-10-12 23:01 发表评论