Algorithm training of Quantum team (20190719)

最新推荐文章于 2020-12-09 10:09:04 发布

原创最新推荐文章于 2020-12-09 10:09:04 发布 · 149 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#多多训练

入门算法记录专栏收录该内容

31 篇文章

订阅专栏

本文介绍了一个有趣的算法挑战，包括如何使用特定字符绘制数字8，并探讨了斐波那契数列的计算方法。通过实例演示，展示了如何处理不同高度的8字图案绘制以及大规模斐波那契数的计算。

Algorithm training of Quantum team (20190719)
A - 画8
谁画8画的好,画的快,今后就发的快,学业发达,事业发达,祝大家发,发,发.
Input
输入的第一行为一个整数N,表示后面有N组数据.
每组数据中有一个字符和一个整数,字符表示画笔,整数(>=5)表示高度.
Output
画横线总是一个字符粗,竖线随着总高度每增长6而增加1个字符宽.当总高度从5增加到6时,其竖线宽度从1增长到2.下圈高度不小于上圈高度,但应尽量接近上圈高度,且下圈的内径呈正方形.
每画一个"8"应空一行,但最前和最后都无空行.
Sample Input
2
A 7
B 8
Sample Output

  AA
AA  AA
AA  AA
  AA
AA  AA
AA  AA
  AA

  BBB
BB   BB
BB   BB
  BBB
BB   BB
BB   BB
BB   BB
  BBB

#include<bits/stdc++.h>
using namespace std;

int main()
{
    int ncase;
    scanf("%d",&ncase);
    while(ncase--)
    {
        char ch ;
        int h;
        scanf(" %c%d",&ch,&h);
        int h3 = max((h-3-(h-3)/2),(h-3)/2);
        int h2 = h - 3 - h3;
        int w = 1 + (h/6);
        for(int i = 1;i<=w;i++)
        {
            printf(" ");
        }
        int tt = h3;
        while(tt--)
            printf("%c",ch);
        printf("\n");
        for(int i =1;i<=h2;i++)
        {
            for(int i =1;i<=2*w+h3;i++)
            {
                if(i>w&&i<=(w + h3))
                    printf(" ");
                else
                    printf("%c",ch);
            }
            printf("\n");
        }
        for(int i = 1;i<=w;i++)
        {
            printf(" ");
        }
        tt = h3;
        while(tt--)
            printf("%c",ch);
        printf("\n");
        for(int i =1;i<=h3;i++)
        {
            for(int i =1;i<=2*w+h3;i++)
            {
                if(i>w&&i<=(w + h3))
                    printf(" ");
                else
                    printf("%c",ch);
            }
            printf("\n");
        }
        for(int i = 1;i<=w;i++)
        {
            printf(" ");
        }
        tt = h3;
        while(tt--)
            printf("%c",ch);
        printf("\n");
        if(ncase)
            printf("\n");
    }
    return 0;
}
/*
2
A 7
B 8
*/

B - Hat’s Fibonacci
A Fibonacci sequence is calculated by adding the previous two members the sequence, with the first two members being both 1.
F(1) = 1, F(2) = 1, F(3) = 1,F(4) = 1, F(n>4) = F(n - 1) + F(n-2) + F(n-3) + F(n-4)
Your task is to take a number as input, and print that Fibonacci number.
Input
Each line will contain an integers. Process to end of file.
Output
For each case, output the result in a line.
Sample Input
100
Sample Output
4203968145672990846840663646

Note:
No generated Fibonacci number in excess of 2005 digits will be in the test data, ie. F(20) = 66526 has 5 digits.

#include<bits/stdc++.h>

using namespace std;
int a[5010][1010];
void init() {
	a[1][0]=1;
	a[2][0]=1;
	for(int i=3; i<=5001; ++i) {
		int c=0;
		for(int j=0; j<=1001; ++j) {
			a[i][j]=(a[i-1][j]+a[i-2][j]+c)%10;
			c=(a[i-1][j]+a[i-2][j]+c)/10;
		}
	}
}
int main() {
	init();
	int n;
	while(~scanf("%d",&n)) {
		int i;
		for(i=1001; i>=0; --i)
			if(a[n][i]!=0)
				break;
		for(; i>=0; --i) {
			printf("%d",a[n][i]);
		}
		printf("\n");
	}
	return 0;
}
/*
100
*/

D - Counting Squares
Your input is a series of rectangles, one per line. Each rectangle is specified as two points(X,Y) that specify the opposite corners of a rectangle. All coordinates will be integers in the range 0 to 100. For example, the line
5 8 7 10
specifies the rectangle who’s corners are(5,8),(7,8),(7,10),(5,10).
If drawn on graph paper, that rectangle would cover four squares. Your job is to count the number of unit(i.e.,1*1) squares that are covered by any one of the rectangles given as input. Any square covered by more than one rectangle should only be counted once.
Input
The input format is a series of lines, each containing 4 integers. Four -1’s are used to separate problems, and four -2’s are used to end the last problem. Otherwise, the numbers are the x-ycoordinates of two points that are opposite corners of a rectangle.
Output
Your output should be the number of squares covered by each set of rectangles. Each number should be printed on a separate line.
Sample Input
5 8 7 10
6 9 7 8
6 8 8 11
-1 -1 -1 -1
0 0 100 100
50 75 12 90
39 42 57 73
-2 -2 -2 -2
Sample Output
8
10000

#include<bits/stdc++.h>

using namespace std;

int main() {
	while(1) {
		bool flag = 0,cf;
		int x1,x2,y1,y2;
		int count = 0;
		int hash[110][110];
		memset(hash,0,sizeof(hash));
		while(cin >> x1 >> y1 >> x2 >> y2) {
			cf = 0;
			if(x1==-1)
				break;

			if(x1==-2) {
				flag=1;
				break;
			}

			int t;
			if(x1>x2) {
				t=x1;
				x1=x2;
				x2=t;
			}
			if(y1>y2) {
				t=y1;
				y1=y2;
				y2=t;
			}
			for(int i=x1; i<x2; i++) //注意是i<x2
				for(int j=y1; j<y2; j++) { //注意是j<y2
					if(hash[i][j]==0) {
						hash[i][j]=1;
						count++;
					}

				}
		}
		cout << count << endl;
		if(flag)
			break;
	}
	return 0;
}
/*
5 8 7 10
6 9 7 8
6 8 8 11
-1 -1 -1 -1
0 0 100 100
50 75 12 90
39 42 57 73
-2 -2 -2 -2
*/

E - Super Jumping! Jumping! Jumping!
Nowadays, a kind of chess game called “Super Jumping! Jumping! Jumping!” is very popular in HDU. Maybe you are a good boy, and know little about this game, so I introduce it to you now.

The game can be played by two or more than two players. It consists of a chessboard（棋盘）and some chessmen（棋子）, and all chessmen are marked by a positive integer or “start” or “end”. The player starts from start-point and must jumps into end-point finally. In the course of jumping, the player will visit the chessmen in the path, but everyone must jumps from one chessman to another absolutely bigger (you can assume start-point is a minimum and end-point is a maximum.). And all players cannot go backwards. One jumping can go from a chessman to next, also can go across many chessmen, and even you can straightly get to end-point from start-point. Of course you get zero point in this situation. A player is a winner if and only if he can get a bigger score according to his jumping solution. Note that your score comes from the sum of value on the chessmen in you jumping path.
Your task is to output the maximum value according to the given chessmen list.
Input
Input contains multiple test cases. Each test case is described in a line as follow:
N value_1 value_2 …value_N
It is guarantied that N is not more than 1000 and all value_i are in the range of 32-int.
A test case starting with 0 terminates the input and this test case is not to be processed.
Output
For each case, print the maximum according to rules, and one line one case.
Sample Input
3 1 3 2
4 1 2 3 4
4 3 3 2 1
0
Sample Output
4
10
3

#include<bits/stdc++.h>

using namespace std;

int main(void){
	
	int n,a[1009],dp[1009];
	 while(scanf("%d",&n)!=EOF&&n){
	 	
	 	for(int i=1;i<=n;i++)
	 	scanf("%d",&a[i]);
	    memset(dp,0,sizeof(dp));
	    dp[1] = a[1];
		 for(int i=1;i<=n;i++){
		 	int flag =0;
		 	for(int j =1;j<i;j++)
		 	   if(a[i] > a[j]){
		 	     dp[i] = max(dp[i],dp[j] + a[i]);
		 	     flag =1;
				}
             if(flag==0)
              dp[i] = a[i];
 		 }
	 	int maxn=0;
		 for(int i=1;i<=n;i++)
		   if(dp[i]>maxn)
		   maxn = dp[i];    	   	
		 
	 	printf("%d\n",maxn);
	 	
	 }
	 
	
	return 0;
}
/*
3 1 3 2
4 1 2 3 4
4 3 3 2 1
0
*/