2014年亚洲区域赛北京赛区现场赛A,D,H,I,K题解(hdu5112,5115,5119,5220,5122)

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下午在HDU上打了一下今年北京区域赛的重现，过了5题，看来单挑只能拿拿铜牌，呜呜。

先将这五题的题解放上来，剩余题目等搞出来再补上

A题

A Curious Matt

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)

Problem Description

There is a curious man called Matt.

One day, Matt's best friend Ted is wandering on the non-negative half of the number line. Matt finds it interesting to know the maximal speed Ted may reach. In order to do so, Matt takes records of Ted’s position. Now Matt has a great deal of records. Please help him to find out the maximal speed Ted may reach, assuming Ted moves with a constant speed between two consecutive records.

 

 

Input

The first line contains only one integer T, which indicates the number of test cases.

For each test case, the first line contains an integer N (2 ≤ N ≤ 10000),indicating the number of records.

Each of the following N lines contains two integers t_i and x_i (0 ≤ t_i, x_i ≤ 10⁶), indicating the time when this record is taken and Ted’s corresponding position. Note that records may be unsorted by time. It’s guaranteed that all t_i would be distinct.

 

 

Output

For each test case, output a single line “Case #x: y”, where x is the case number (starting from 1), and y is the maximal speed Ted may reach. The result should be rounded to two decimal places.

 

 

Sample Input

2

3

2 2

1 1

3 4

3

0 3

1 5

2 0

Sample Output

Case #1: 2.00

Case #2: 5.00

Hint

In the first sample, Ted moves from 2 to 4 in 1 time unit. The speed 2/1 is maximal. In the second sample, Ted moves from 5 to 0 in 1 time unit. The speed 5/1 is maximal.

题目要求求出最大的速度

分析：以时间为key值，排序，然后遍历一遍

#include <iostream>
#include <cstring>
#include <iomanip>
#include <cmath>
#include <algorithm>
using namespace std;
int a[10010];
int b[10010];
double c[10010];
int id[10010];
bool cmp(int x,int y)
{
    return a[x]<=a[y];
}
int main()
{
    ios::sync_with_stdio(false);
    int t;
    int n;
    int cas=1;
    cin>>t;
    while(t--)
    {
        cin>>n;
        a[0]=0,b[0]=0;
        for(int i=0;i<n;i++)
        {
            cin>>a[i]>>b[i];
            id[i]=i;
        }
        sort(id,id+n,cmp);
        double maxx=0;
        for(int i=1;i<n;i++)
        {
            maxx=max(maxx,fabs((b[id[i]]-b[id[i-1]])*1.0/((a[id[i]]-a[id[i-1]])*1.0)));
        }
        cout<<"Case #"<<cas++<<": ";
        cout<<fixed<<setprecision(2)<<maxx<<endl;
    }
    return 0;
}

D题

Dire Wolf

Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)

Problem Description

Dire wolves, also known as Dark wolves, are extraordinarily large and powerful wolves. Many, if not all, Dire Wolves appear to originate from Draenor.
Dire wolves look like normal wolves, but these creatures are of nearly twice the size. These powerful beasts, 8 - 9 feet long and weighing 600 - 800 pounds, are the most well-known orc mounts. As tall as a man, these great wolves have long tusked jaws that look like they could snap an iron bar. They have burning red eyes. Dire wolves are mottled gray or black in color. Dire wolves thrive in the northern regions of Kalimdor and in Mulgore.
Dire wolves are efficient pack hunters that kill anything they catch. They prefer to attack in packs, surrounding and flanking a foe when they can.
— Wowpedia, Your wiki guide to the World of Warcra

Matt, an adventurer from the Eastern Kingdoms, meets a pack of dire wolves. There are N wolves standing in a row (numbered with 1 to N from left to right). Matt has to defeat all of them to survive.

Once Matt defeats a dire wolf, he will take some damage which is equal to the wolf’s current attack. As gregarious beasts, each dire wolf i can increase its adjacent wolves’ attack by b_i. Thus, each dire wolf i’s current attack consists of two parts, its basic attack ai and the extra attack provided by the current adjacent wolves. The increase of attack is temporary. Once a wolf is defeated, its adjacent wolves will no longer get extra attack from it. However, these two wolves (if exist) will become adjacent to each other now.

For example, suppose there are 3 dire wolves standing in a row, whose basic attacks ai are (3, 5, 7), respectively. The extra attacks b_i they can provide are (8, 2, 0). Thus, the current attacks of them are (5, 13, 9). If Matt defeats the second wolf first, he will get 13 points of damage and the alive wolves’ current attacks become (3, 15).

As an alert and resourceful adventurer, Matt can decide the order of the dire wolves he defeats. Therefore, he wants to know the least damage he has to take to defeat all the wolves.

 

 

Input

The first line contains only one integer T , which indicates the number of test cases. For each test case, the first line contains only one integer N (2 ≤ N ≤ 200).

The second line contains N integers a_i (0 ≤ a_i ≤ 100000), denoting the basic attack of each dire wolf.

The third line contains N integers b_i (0 ≤ b_i ≤ 50000), denoting the extra attack each dire wolf can provide.

 

 

Output

For each test case, output a single line “Case #x: y”, where x is the case number (starting from 1), y is the least damage Matt needs to take.

 

 

Sample Input

2

3

3 5 7

8 2 0

10

1 3 5 7 9 2 4 6 8 10

9 4 1 2 1 2 1 4 5 1

 

 

Sample Output

Case #1: 17

Case #2: 74

Hint

In the first sample, Matt defeats the dire wolves from left to right. He takes 5 + 5 + 7 = 17 points of damage which is the least damage he has to take.

 题意：有n头狼排成一排，每只狼都对相邻的狼的攻击力有加成作用，每杀死一只狼所受到的伤害为当前狼的攻击力（算上加成的部分），被杀死之后的狼对相邻的狼的攻击力的加成会被取消，同时，原先与被杀死的狼相邻的两头狼会变成相邻的狼。要求使得受到的伤害值最小，求出最小值。

分析：区间dp水题

dp[i][j]表示从第i头狼到第j头狼全部被杀死所受到的最小伤害。a[i]表示第i头狼的初始攻击力，b[i]表示第i头狼对相邻狼的加成值。

dp[i][j]=min(dp[i][k-1]+a[k]+dp[k+1][j])+b[i-1]+b[j+1]; (i<k<j&&j>i)

dp[i][j]=a[i]+b[i-1]+b[j+1];　　(i=j)

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
int a[310],b[310];
int dp[310][310];
int main()
{
    ios::sync_with_stdio(false);
    int t,n;
    cin>>t;
    int cas=1;
    while(t--)
    {
        cin>>n;
        for(int i=1;i<=n;i++)cin>>a[i];
        for(int i=1;i<=n;i++)cin>>b[i];
        b[0]=b[n+1]=0;
        for(int p=0;p<n;p++)
        {
            for(int i=1;i<=n;i++)
            {
                int j=i+p;
                if(i==j)
                {
                    dp[i][j]=a[i]+b[i-1]+b[i+1];
                    continue;
                }
                if(j>n)break;
                dp[i][j]=min(a[i]+dp[i+1][j],a[j]+dp[i][j-1]);
                int temp=0;
                for(int k=i+1;k<j;k++)
                {
                    dp[i][j]=min(dp[i][j],dp[i][k-1]+a[k]+dp[k+1][j]);
                }
                dp[i][j]+=b[i-1]+b[j+1];
            }
        }
        cout<<"Case #"<<cas++<<": ";
        cout<<dp[1][n]<<endl;
    }
    return 0;
}

H题

Happy Matt Friends

Time Limit: 6000/6000 MS (Java/Others)    Memory Limit: 510000/510000 K (Java/Others)

Problem Description

Matt has N friends. They are playing a game together.

Each of Matt’s friends has a magic number. In the game, Matt selects some (could be zero) of his friends. If the xor (exclusive-or) sum of the selected friends’magic numbers is no less than M , Matt wins.

Matt wants to know the number of ways to win.

 

 

Input

The first line contains only one integer T , which indicates the number of test cases.

For each test case, the first line contains two integers N, M (1 ≤ N ≤ 40, 0 ≤ M ≤ 10⁶).

In the second line, there are N integers ki (0 ≤ k_i ≤ 10⁶), indicating the i-th friend’s magic number.

 

 

Output

For each test case, output a single line “Case #x: y”, where x is the case number (starting from 1) and y indicates the number of ways where Matt can win.

 

 

Sample Input

2

3 2

1 2 3

3 3

1 2 3

 

 

Sample Output

Case #1: 4

Case #2: 2

Hint

In the first sample, Matt can win by selecting: friend with number 1 and friend with number 2. The xor sum is 3. friend with number 1 and friend with number 3. The xor sum is 2. friend with number 2. The xor sum is 2. friend with number 3. The xor sum is 3. Hence, the answer is 4.

题意：求从n个数中挑出一些数，并且这些数的异或和不小与M的方案数，挑出的数的数目可为0.

分析：dp

dp[i][j]表示从前i个数中挑出一些数，并且这些数的异或和为j的方案数

鉴于直接数组开不下，用一下滚动数组即可

#include <iostream>
#include <cstring>
using namespace std;
int a[110];
long long dp[2][(1<<20)+10];
int main()
{
    ios::sync_with_stdio(false);
    int t;
    cin>>t;
    int cas=1;
    while(t--)
    {
        int n,m;
        cin>>n>>m;
        memset(dp,0,sizeof(dp));
        for(int i=0;i<n;i++)cin>>a[i];
        cout<<"Case #"<<cas++<<": ";
        dp[1][0]=1;
        int k;
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<(1<<20);j++)dp[i&1][j]=dp[!(i&1)][j];
            for(int j=0;j<(1<<20);j++)
            {
                k=a[i]^j;
                dp[i&1][k]+=dp[!(i&1)][j];
            }
        }
        long long ans=0;
        for(int i=m;i<(1<<20);i++)ans+=dp[!(n&1)][i];
        cout<<ans<<endl;
    }
    return 0;
}

I题

Intersection

Time Limit: 4000/4000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)

Problem Description

Matt is a big fan of logo design. Recently he falls in love with logo made up by rings. The following figures are some famous examples you may know.


A ring is a 2-D figure bounded by two circles sharing the common center. The radius for these circles are denoted by r and R (r < R). For more details, refer to the gray part in the illustration below.


Matt just designed a new logo consisting of two rings with the same size in the 2-D plane. For his interests, Matt would like to know the area of the intersection of these two rings.

 

Input

The first line contains only one integer T (T ≤ 10⁵), which indicates the number of test cases. For each test case, the first line contains two integers r, R (0 ≤ r < R ≤ 10).

Each of the following two lines contains two integers x_i, y_i (0 ≤ x_i, y_i ≤ 20) indicating the coordinates of the center of each ring.

 

Output

For each test case, output a single line “Case #x: y”, where x is the case number (starting from 1) and y is the area of intersection rounded to 6 decimal places.

 

Sample Input

2

2 3

0 0

0 0

2 3

0 0

5 0

 

Sample Output

Case #1: 15.707963

Case #2: 2.250778

 

题意：给出两个全等圆环，求两圆环的面积的交

分析：根据容斥，直接推出公式，转化成求两圆面积交的问题

ans=大圆的面积交+小圆的面积交-大圆1与小圆2的面积交-大圆2与小圆1的面积交

#include <iostream>
#include <cstring>
#include <iomanip>
#include <cmath>
#include <algorithm>
#include <cstdio>
using namespace std;

#define PI acos(-1.0)
double area(double x1,double y1,double r1,double x2,double y2,double r2)
{
    double d=(x1-x2)*(x1-x2)+(y1-y2)*(y1-y2);
    if(d>=(r1+r2)*(r1+r2))return 0;
    if(d<=(r1-r2)*(r1-r2))return r1<r2 ? PI*r1*r1 : PI*r2*r2;
    d=sqrt(d);
    double a1=acos((r1*r1+d*d-r2*r2)/(2.0*r1*d));
    double a2=acos((r2*r2+d*d-r1*r1)/(2.0*r2*d));
    double s1=a1*r1*r1;
    double s2=a2*r2*r2;
    double t=(r1+r2+d)/2.0;
    t=2.0*sqrt(t*(t-r1)*(t-r2)*(t-d));
    return s1+s2-t;
}
int main()
{
    //ios::sync_with_stdio(false);
    int t;
    double r1,r2,x1,y1,x2,y2;
    scanf("%d",&t);
    int cas=1;
    while(t--)
    {
        scanf("%lf%lf%lf%lf%lf%lf",&r1,&r2,&x1,&y1,&x2,&y2);
        double ans=0;
        ans-=area(x1,y1,r2,x2,y2,r1)*2;
        ans+=area(x1,y1,r2,x2,y2,r2);
        ans+=area(x1,y1,r1,x2,y2,r1);
        printf("Case #%d: ",cas++);
        printf("%.6lf\n",ans);
    }
    return 0;
}

K题

K.Bro Sorting

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)

Problem Description

Matt’s friend K.Bro is an ACMer.

Yesterday, K.Bro learnt an algorithm: Bubble sort. Bubble sort will compare each pair of adjacent items and swap them if they are in the wrong order. The process repeats until no swap is needed.

Today, K.Bro comes up with a new algorithm and names it K.Bro Sorting.

There are many rounds in K.Bro Sorting. For each round, K.Bro chooses a number, and keeps swapping it with its next number while the next number is less than it. For example, if the sequence is “1 4 3 2 5”, and K.Bro chooses “4”, he will get “1 3 2 4 5” after this round. K.Bro Sorting is similar to Bubble sort, but it’s a randomized algorithm because K.Bro will choose a random number at the beginning of each round. K.Bro wants to know that, for a given sequence, how many rounds are needed to sort this sequence in the best situation. In other words, you should answer the minimal number of rounds needed to sort the sequence into ascending order. To simplify the problem, K.Bro promises that the sequence is a permutation of 1, 2, . . . , N .

 

Input

The first line contains only one integer T (T ≤ 200), which indicates the number of test cases. For each test case, the first line contains an integer N (1 ≤ N ≤ 10⁶).

The second line contains N integers a_i (1 ≤ a_i ≤ N ), denoting the sequence K.Bro gives you.

The sum of N in all test cases would not exceed 3 × 10⁶.

 

Output

For each test case, output a single line “Case #x: y”, where x is the case number (starting from 1), y is the minimal number of rounds needed to sort the sequence.

 

Sample Input

2

5

5 4 3 2 1

5

5 1 2 3 4

 

Sample Output

Case #1: 4

Case #2: 1

Hint

In the second sample, we choose “5” so that after the first round, sequence becomes “1 2 3 4 5”, and the algorithm completes.

题意：求题目完成排序需要题目所述的最小的round次数

分析：每次把不符合排序的最大的数进行swap，那么，这个数在经过一个round之后，所有大于等于它的数一定是最终的排列。

由此，可以将问题转化为判断一个数的右边是否有必该数小的数，若有，则需要一次round。

对于此问题，只需要从右往左统计，并且不断更新最小值，若当前数为最小值，则将最小值更新为当前数，否则round+1

#include <iostream>
#include <cstdio>
using namespace std;
int a[1000010];
int main()
{
    ios::sync_with_stdio(false);
    int n;
    int t;
    int cas=1;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        int k,s=1,m=0;
        int ans=0;
        int minn=1e7;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        for(int i=n;i>0;i--)
        {
            if(a[i]<minn)minn=a[i];
            else ans++;
        }
        printf("Case #%d: %d\n",cas++,ans);

    }
    return 0;
}