2016ACM青島區域賽題解

A、Relic Discovery_hdu5982

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 57    Accepted Submission(s): 49

Problem Description

Recently, paleoanthropologists have found historical remains on an island in the Atlantic Ocean. The most inspiring thing is that they excavated in a magnificent cave and found that it was a huge tomb. Inside the construction,researchers identified a large number of skeletons, and funeral objects including stone axe, livestock bones and murals. Now, all items have been sorted, and they can be divided into N types. After they were checked attentively, you are told that there are Ai items of the i-th type. Further more, each item of the i-th type requires Bi million dollars for transportation, analysis, and preservation averagely. As your job, you need to calculate the total expenditure.

 

Input

The first line of input contains an integer T which is the number of test cases. For each test case, the first line contains an integer N which is the number of types. In the next N lines, the i-th line contains two numbers  Ai and Bi as described above. All numbers are positive integers and less than 101.

 

Output

For each case, output one integer, the total expenditure in million dollars.

 

Sample Input

1 2 1 2 3 4

 

Sample Output

14

 

Source

2016ACM/ICPC亞洲區青島站-重現賽（感謝中國石油大學）

 

 

Recommend

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思路：第一題很簡單，求費用，把每次的乘積都加起來就行了。

#include <iostream>
#include <cstdio>

using namespace std;

int main()
{
    int n;
    int t;
    int a,b;
    scanf("%d",&t);
    for(int i=0;i<t;i++){
        scanf("%d",&n);
        int sum=0;
        for(int j=0;j<n;j++){
            scanf("%d %d",&a,&b);
            sum+=a*b;
        }
        printf("%d\n",sum);
    }

    return 0;
}

 

B、Pocket Cube_5983

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 23    Accepted Submission(s): 9

Problem Description

The Pocket Cube, also known as the Mini Cube or the Ice Cube, is the 2 × 2 × 2 equivalence of a Rubik’s Cube.
The cube consists of 8 pieces, all corners.
Each piece is labeled by a three dimensional coordinate (h, k, l) where h, k, l ∈ {0, 1}. Each of the six faces owns four small faces filled with a positive integer.
For each step, you can choose a certain face and turn the face ninety degrees clockwise or counterclockwise.
You should judge that if one can restore the pocket cube in one step. We say a pocket cube has been restored if each face owns four same integers.

 

 

Input

The first line of input contains one integer N(N ≤ 30) which is the number of test cases.
For each test case, the first line describes the top face of the pocket cube, which is the common 2 × 2 face of pieces
labelled by (0, 0, 1),(0, 1, 1),(1, 0, 1),(1, 1, 1). Four integers are given corresponding to the above pieces.
The second line describes the front face, the common face of (1, 0, 1),(1, 1, 1),(1, 0, 0),(1, 1, 0). Four integers are
given corresponding to the above pieces.
The third line describes the bottom face, the common face of (1, 0, 0),(1, 1, 0),(0, 0, 0),(0, 1, 0). Four integers are
given corresponding to the above pieces.
The fourth line describes the back face, the common face of (0, 0, 0),(0, 1, 0),(0, 0, 1),(0, 1, 1). Four integers are
given corresponding to the above pieces.
The fifth line describes the left face, the common face of (0, 0, 0),(0, 0, 1),(1, 0, 0),(1, 0, 1). Four integers are given
corresponding to the above pieces.
The six line describes the right face, the common face of (0, 1, 1),(0, 1, 0),(1, 1, 1),(1, 1, 0). Four integers are given
corresponding to the above pieces.
In other words, each test case contains 24 integers a, b, c to x. You can flat the surface to get the surface development
as follows.

+ - + - + - + - + - + - +
| q | r | a | b | u | v |
+ - + - + - + - + - + - +
| s | t | c | d | w | x |
+ - + - + - + - + - + - +
        | e | f |
        + - + - +
        | g | h |
        + - + - +
        | i | j |
        + - + - +
        | k | l |
        + - + - +
        | m | n |
        + - + - +
        | o | p |
        + - + - +

Output

For each test case, output YES if can be restored in one step, otherwise output NO.

 

Sample Input

4 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 5 5 5 5 6 6 6 6 6 6 6 6 1 1 1 1 2 2 2 2 3 3 3 3 5 5 5 5 4 4 4 4 1 4 1 4 2 1 2 1 3 2 3 2 4 3 4 3 5 5 5 5 6 6 6 6 1 3 1 3 2 4 2 4 3 1 3 1 4 2 4 2 5 5 5 5 6 6 6 6

 

Sample Output

YES YES YES NO

 

Source

2016ACM/ICPC亞洲區青島站-重現賽（感謝中國石油大學）

 

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jiangzijing2015   |   We have carefully selected several similar problems for you:   5994  5993  5992  5991  5990 

思路：在比賽的時候理解錯意思了，以為是判斷能不能復原模仿，而沒看到只需要一步。

魔方是2*2的，有三種轉的方法，上面順時針轉90度，逆時針轉，前面順時針，前面逆時針，左邊順時針（相當於右邊順時針），左邊逆時針（相當於左邊逆時針）

還有再加上不操作算是一種，總共七種類，模擬出來就行了。

#include <iostream>
#include <cstdio>

using namespace std;

int a[7][5];
int b[7][5];

bool judge(){
    int i=0;
    for(int j=1;j<=6;j++){
        if(a[j][1]==a[j][2]&&a[j][2]==a[j][3]&&a[j][3]==a[j][4]){
            i++;
        }
    }
    if(i==6){
        return true;
    }else{
        return false;
    }
}

void opera1(){//前面順時針
    int t1,t2;
    t1=a[6][3];
    t2=a[6][4];

    a[6][3]=a[1][3];
    a[6][4]=a[1][4];

    a[1][3]=a[5][3];
    a[1][4]=a[5][4];

    a[5][3]=a[3][2];
    a[5][4]=a[3][1];

    a[3][1]=t2;
    a[3][2]=t1;
}
void opera2(){//前面逆時針
    int t1,t2;
    t1=a[5][3];
    t2=a[5][4];

    a[5][3]=a[1][3];
    a[5][4]=a[1][4];

    a[1][3]=a[6][3];
    a[1][4]=a[6][4];

    a[6][3]=a[3][2];
    a[6][4]=a[3][1];

    a[3][1]=t2;
    a[3][2]=t1;
}
void opera3(){//上面順時針
    int t1,t2;
    t1=a[2][1];
    t2=a[2][2];

    a[2][1]=a[6][3];
    a[2][2]=a[6][1];

    a[6][3]=a[4][4];
    a[6][1]=a[4][3];

    a[4][4]=a[5][2];
    a[4][3]=a[5][4];

    a[5][2]=t1;
    a[5][4]=t2;
}
void opera4(){//上面逆時針
    int t1,t2;
    t1=a[2][1];
    t2=a[2][2];

    a[2][1]=a[5][2];
    a[2][2]=a[5][4];

    a[5][2]=a[4][4];
    a[5][4]=a[4][3];

    a[4][4]=a[6][3];
    a[4][3]=a[6][1];

    a[6][1]=t2;
    a[6][3]=t1;
}
void opera5(){//左邊順指針
    int t1,t2;
    t1=a[1][1];
    t2=a[1][3];

    a[1][1]=a[4][1];
    a[1][3]=a[4][3];

    a[4][1]=a[3][1];
    a[4][3]=a[3][3];

    a[3][1]=a[2][1];
    a[3][3]=a[2][3];

    a[2][1]=t1;
    a[2][3]=t2;
}
void opera6(){//左邊逆時針
    int t1,t2;
    t1=a[1][1];
    t2=a[1][3];

    a[1][1]=a[2][1];
    a[1][3]=a[2][3];

    a[2][1]=a[3][1];
    a[2][3]=a[3][3];

    a[3][1]=a[4][1];
    a[3][3]=a[4][3];

    a[4][1]=t1;
    a[4][3]=t2;
}
int main()
{
    int n;
    scanf("%d",&n);
    while(n--){
        for(int i=1;i<=6;i++){
            for(int j=1;j<=4;j++){
                scanf("%d",&a[i][j]);
                b[i][j]=a[i][j];
            }
        }
        if(judge()){
            printf("YES\n");
            continue;
        }
        opera1();
        if(judge()){
            printf("YES\n");
            continue;
        }
        for(int i=1;i<=6;i++){
            for(int j=1;j<=4;j++){
                a[i][j]=b[i][j];
            }
        }
        opera2();
        if(judge()){
            printf("YES\n");
            continue;
        }
        for(int i=1;i<=6;i++){
            for(int j=1;j<=4;j++){
                a[i][j]=b[i][j];
            }
        }
        opera3();
        if(judge()){
            printf("YES\n");
            continue;
        }
        for(int i=1;i<=6;i++){
            for(int j=1;j<=4;j++){
                a[i][j]=b[i][j];
            }
        }
        opera4();
        if(judge()){
            printf("YES\n");
            continue;
        }
        for(int i=1;i<=6;i++){
            for(int j=1;j<=4;j++){
                a[i][j]=b[i][j];
            }
        }
        opera5();
        if(judge()){
            printf("YES\n");
            continue;
        }
        for(int i=1;i<=6;i++){
            for(int j=1;j<=4;j++){
                a[i][j]=b[i][j];
            }
        }
        opera6();
        if(judge()){
            printf("YES\n");
            continue;
        }
        for(int i=1;i<=6;i++){
            for(int j=1;j<=4;j++){
                a[i][j]=b[i][j];
            }
        }
        printf("NO\n");
    }
    return 0;
}

 

C、Pocky_hdu5984

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 54    Accepted Submission(s): 20

Problem Description

Let’s talking about something of eating a pocky. Here is a Decorer Pocky, with colorful decorative stripes in the coating, of length L.
While the length of remaining pocky is longer than d, we perform the following procedure. We break the pocky at any point on it in an equal possibility and this will divide the remaining pocky into two parts. Take the left part and eat it. When it is not longer than d, we do not repeat this procedure.
Now we want to know the expected number of times we should repeat the procedure above. Round it to 6 decimal places behind the decimal point.

 

Input

The first line of input contains an integer N which is the number of test cases. Each of the N lines contains two float-numbers L and d respectively with at most 5 decimal places behind the decimal point where 1 ≤ d, L ≤ 150.

 

Output

For each test case, output the expected number of times rounded to 6 decimal places behind the decimal point in a line.

 

Sample Input

6 1.0 1.0 2.0 1.0 4.0 1.0 8.0 1.0 16.0 1.0 7.00 3.00

 

Sample Output

0.000000 1.693147 2.386294 3.079442 3.772589 1.847298

 

Source

2016ACM/ICPC亞洲區青島站-重現賽（感謝中國石油大學）

 

Recommend

jiangzijing2015   |   We have carefully selected several similar problems for you:   5994  5993  5992  5991  5990 

思路：說實話比賽的時候根本不知道怎么做出來的。

好久之后才隊友想起來，輸出了一下log2，突然發現了規律。

賽后看別的隊，一看0.693147就知道log2了，這就是做題多了有經驗了啊。

但當時說的題解好像是求微積分，還是求導呢，把公式推導出來的。。。

#include <iostream>
#include <cstdio>
#include <cmath>

using namespace std;

int main()
{
    int n;
    double a,b;
    scanf("%d",&n);
    while(n--){
        scanf("%lf%lf",&a,&b);
        if(a<=b){
            printf("0.000000\n");
            continue;
        }
        printf("%.6lf\n",log(a)-log(b)+1);
    }
    return 0;
}