[旧]メモり。メモる。(お引越ししました→http://hatone.hateblo.jp/)

2012-05-22 引越しのお知らせ

id:hatoneが欲しいid:hatyoneです。

はてなブログの方へお引越ししました。


メモりメモる

2010-06-11 バナナの謎はまだ謎なのだぞ

#include<iostream>
#include<math.h>

#define M_PI       3.14159265358979323846

using namespace std;


int main(){
	int ans =0;
	int n=0;
	int m=0;


	while(1){
		cin >> n;
		cout<<n;

		if(n==0)break;
		int oreA[10][2];
		// answer
		cin >> m;
		for(int j=0; j<m; j++)
		{
			cin >>oreA[j][0];
			cin >>oreA[j][1];
		}

		// hoka
		for(int i=0; i<n-1; i++)
		{
			cin >> m;
			int dare[10][2];
			for(int j=0; j<m; j++)
			{
				cin >> dare[j][0];
				cin >> dare[j][1];

				cout >> dare[j][0];
				cout >> dare[j][1];
			}

			//for(int h=0; h <4 ; h++)
			//{
			//	int c=0;
			//	for(int j=0; j<m; j++)
			//	{
			//		int tempx=dare[j][0];
			//		dare[j][0]=dare[j][0]*cos(M_PI/2)+dare[j][1]*sin(M_PI/2);
			//		dare[j][1]=tempx*sin(M_PI/2)*(-1)+dare[j][0]*cos(M_PI/2);

			//		if(oreA[j][0]==dare[j][0] && oreA[j][1]==dare[j][1]){
			//			c++;
			//		}
			//	}
			//	if(c==m){
			//		cout << i+1;
			//	}

			//}





		}


	}

	return 0;
}

2010-02-14

[][]16問目

未確認。あんさいんどろんぐろんぐなので後でとく。

#include<iostream>

using namespace std;

int main(){

  unsigned long long int p=1;
  unsigned long long int n=0;

  for(int i=0;i<1000;i++)
    {
      p*=2;
    }
  cout<<p<<endl;

  while(p>0)
    {
      n+= p%10;
      p/=10;
      

      cout<<p<<endl;
      
    }

  cout<<"n is "<<n<<endl;

}

[][]14問目

#include<iostream>
using namespace std;

int main(){

  unsigned long int max=0;
  unsigned long int maxans=0;
  unsigned long int p=0;
  unsigned long int count=0;

  for(unsigned long int i=999999;i>=0;i--)
   {
      count=0;
      p=i;
      while(p!=1)	
	{
	  if(p%2==0)
	    {
	      p=p/2;
	      count++;     
	    }
	  else
	    {
	      p=3*p+1;
	      count++;
	    }
	}
      
      //  cout<<count<<endl;
      if(max<count)
	{
	maxans=i;
	max=count;
	cout<<maxans<<endl;
	}

      if(i==1)break;
       }

  cout<<"ans is"<<maxans<<endl;

  return 0;


}

[][]13問目

#include<iostream>
using namespace std;

int main(){

  int num[100][50];
  int buf[50];
  int mod[50];
  int ans[50];
  char c;
  
  for(int i=0;i<100;i++)
    {
      for(int p=0;p<50;p++)
	{
	  cin>>c;
	  num[i][p]=c-'0';

	  buf[p]=0;
	  mod[p]=0;
	  ans[p]=0;
	}
    }

  for(int i=0;i<50;i++)
    {
      for(int p=0;p<100;p++)
	{
	  mod[49-i]+=num[p][49-i];
	}

      mod[49-i]+=buf[49-i];
      ans[49-i]=mod[49-i]%10;
      buf[48-i]=mod[49-i]/10;
    }
  
  cout<<"ans"<<mod[0]<<endl;

  for(int i=1;i<50;i++)
    {
      cout<<i<<":"<<ans[i]<<endl;
    }
  
 


  return 0;


}

2010-02-12

[][]12問目

#include <iostream>
#include <math.h>
using namespace std;
int main(){	
	int sq=0;
	int count=0;
	int prv=0;

	for(long long int i=1;i<100000000000000000;i++)
	{
		
		sq+=i;
		if( sqrt((double)sq) - floor(sqrt((double)sq))==0.0) count--;

		for(int n=1; n<=sq; n++ )
		{	

			
			if(sq%n==0 && n > prv && n!=1)break;
			if(sq%n==0)
			{		
				prv=sq/n;
				count+=2;
			}		
		}

		cout<<count<<" sq"<<sq<<endl;

		if(count >501)
		{
			break;
		}

		count=0;
	}

	cout<<"end sq"<<sq<<"  count"<<count<<endl;	

}

[][]11問目

斜めって2方向あるんですよ奥さん。

In the 2020 grid below, four numbers along a diagonal line have been marked in red.

08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08

49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00

81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65

52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91

22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80

24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50

32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70

67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21

24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72

21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95

78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92

16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57

86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58

19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40

04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66

88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69

04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36

20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16

20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54

01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48

The product of these numbers is 26 63 78 14 = 1788696.

What is the greatest product of four adjacent numbers in any direction (up, down, left, right, or diagonally) in the 2020 grid?

#include <iostream>
#include <math.h>
using namespace std;
int main(){	

int num[20][20];
int max=0;

for(int i=0;i<20;i++)
{
	for(int n=0;n<20;n++)
	{
	
		cin>>num[i][n];
	}
}



for(int i=0;i<20;i++)
{
	for(int n=0;n<20;n++)
	{
		
		if(n>=3)
		{
			if(max< num[i][n-3]*num[i][n-2]*num[i][n-1]*num[i][n])max=num[i][n-3]*num[i][n-2]*num[i][n-1]*num[i][n];
			if(max< num[n-3][i]*num[n-2][i]*num[n-1][i]*num[n][i])max=num[n-3][i]*num[n-2][i]*num[n-1][i]*num[n][i];
		}

		if(n>=3&&i>=3)
		{
			if(max < num[i-3][n-3]*num[i-2][n-2]*num[i-1][n-1]*num[i][n])max=num[i-3][n-3]*num[i-2][n-2]*num[i-1][n-1]*num[i][n];
			if(max < num[i-3][n]*num[i-2][n-1]*num[i-1][n-2]*num[i][n-3])max=num[i-3][n]*num[i-2][n-1]*num[i-1][n-2]*num[i][n-3];
		}

		cout<<max<<endl;	
			
	}

	
	}

cout<<"this is ans:"<<max<<endl;

}

[][]第10問

The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.

Find the sum of all the primes below two million.

#include <iostream>
#include <math.h>
using namespace std;



long long int prime[1000000];
long long int table[100000000];
int primeCall(){


	int flag=0;
	
	prime[0]=2;
	prime[1]=3;


	for(long long int i=0;i<3000000;i++)
	{
			if(i%2==0)table[i]=0;
			else table[i]=1;
	}

	
	for(long long int p=1;p<1000000;p++)
	{
		if(p%100==0)cout<<prime[p]<<endl;
		for(long long int i=prime[p];i<2001000;i+=2)
		{
			if(i%prime[p]!=0 && flag==0 && table[i]==1)
			{
				prime[p+1]=i;
				flag=1;

			}
			if(i%prime[p]==0)
			{
				table[i]=0;
			}
	
		}


		flag=0;

		if(prime[p+1]>2000000)break;

	}



2010-01-24

[][]8問目。

Find the greatest product of five consecutive digits in the 1000-digit number.

73167176531330624919225119674426574742355349194934

96983520312774506326239578318016984801869478851843

85861560789112949495459501737958331952853208805511

12540698747158523863050715693290963295227443043557

66896648950445244523161731856403098711121722383113

62229893423380308135336276614282806444486645238749

30358907296290491560440772390713810515859307960866

70172427121883998797908792274921901699720888093776

65727333001053367881220235421809751254540594752243

52584907711670556013604839586446706324415722155397

53697817977846174064955149290862569321978468622482

83972241375657056057490261407972968652414535100474

82166370484403199890008895243450658541227588666881

16427171479924442928230863465674813919123162824586

17866458359124566529476545682848912883142607690042

24219022671055626321111109370544217506941658960408

07198403850962455444362981230987879927244284909188

84580156166097919133875499200524063689912560717606

05886116467109405077541002256983155200055935729725

71636269561882670428252483600823257530420752963450

#include<iostream>

using namespace std;
char str[]="7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450";

int main(){	
	int i=0;//1000桁目、100桁目、10桁目、1桁目をカウントする
	int num=1;
	int max=0;

	for(int p=0;str[p+4]!='\0';p++) {
		for(int m=10000; m>=1; m/=10,i++) 
			{
				cout<<(str[i+p]-'0');
				num*=(str[i+p]-'0');
			}
		cout<<endl;
		cout<<num<<endl;
		if(max < num) max=num; 
		i=0;
		num=1;
	}
	cout<<max<<endl;
}

[][]7問目。

By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.

What is the 10001st prime number?

#include <iostream>

using namespace std;


long long  int primeCall(unsigned int NM){

	long long int prime[50000];
	long long int *box=new long long int[1000000];

	long long int p=0;

	for(p=0;p<NM;p++)prime[p]=0;
	prime[0]=2;
	box[0]=0;

	for(p=1;p<=NM;p++)
	{	
		for(long long int i=prime[p-1];i<1000000;i++)
		{
			if((i%prime[p-1])==0)box[i]=0;
			else if (box[i]!=0)box[i]=i;			
		}
		
		for(long long int i=2;i<1000000;i++)
		{
			if(box[i]!=0)
			{
				prime[p]=i;
				break;
			}
		}
	}

	delete [] box;

	return prime[NM-1];
}
int main()
{
	cout<<primeCall(10001)<<endl;
}

[][]6問目。

The sum of the squares of the first ten natural numbers is,

1^2 + 2^2 + ... + 10^2 = 385

The square of the sum of the first ten natural numbers is,

(1 + 2 + ... + 10)^2 = 552 = 3025

Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 385 = 2640.

Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.

#include <iostream>

using namespace std;

int main()
{
	
	unsigned long long int sum=0;
	unsigned long long int sq=0;
	for(int i=1;i<=100;i++)
	{
		sum+=(i*i);

		sq+=i;
	}
	cout<<(sq*sq)-sum<<endl;
}

[][]5問目。


2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.

What is the smallest number that is evenly divisible by all of the numbers from 1 to 20?

#include <iostream>

using namespace std;

int main()
{
	
	for(int i=2520;i<=1000000000;i++)
	{
		
		if( (i%11)==0 && (i%12)==0 && (i%13)==0 && (i%14)==0 && (i%15)==0 && (i%16)==0 && (i%17)==0 && (i%18)==0 && (i%19)==0 &&(i%20)==0 )cout<<i<<endl;

	}
}

[][]4問目。

A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91*99.

Find the largest palindrome made from the product of two 3-digit numbers.