projecteuler.net中的题基本都和数字相关，非常容易上手。现在找出几个有趣的题，分享一下。
1.
Problem 8</h2>
11 January 2002</div>
Find the greatest product of five consecutive digits in the 1000-digit number.</p> 73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450
找出5个连续的数的最大乘积
我正在学C，这是代码。
#include
int main()
{
char ch[1000]=
"73167176531330624919225119674426574742355349194934"
"96983520312774506326239578318016984801869478851843"
"85861560789112949495459501737958331952853208805511"
"12540698747158523863050715693290963295227443043557"
"66896648950445244523161731856403098711121722383113"
"62229893423380308135336276614282806444486645238749"
"30358907296290491560440772390713810515859307960866"
"70172427121883998797908792274921901699720888093776"
"65727333001053367881220235421809751254540594752243"
"52584907711670556013604839586446706324415722155397"
"53697817977846174064955149290862569321978468622482"
"83972241375657056057490261407972968652414535100474"
"82166370484403199890008895243450658541227588666881"
"16427171479924442928230863465674813919123162824586"
"17866458359124566529476545682848912883142607690042"
"24219022671055626321111109370544217506941658960408"
"07198403850962455444362981230987879927244284909188"
"84580156166097919133875499200524063689912560717606"
"05886116467109405077541002256983155200055935729725"
"71636269561882670428252483600823257530420752963450";
int i,temp,product=7*3*1*6*7;
for(i=0;i<996;i++)
{
temp=(ch[i]-'0')*(ch[i+1]-'0')*(ch[i+2]-'0')*(ch[i+3]-'0')*(ch[i+4]-'0');
if(temp>product)
product=temp;
}
printf("%dn",product);
return 0;
}
2.
Problem 11</h2>
22 February 2002</div>
In the 20
20 grid below, four numbers along a diagonal line have been marked in red.</p>
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</b></span> 38 40 67 59 54 70 66 18 38 64 70
67 26 20 68 02 62 12 20 95 63</b></span> 94 39 63 08 40 91 66 49 94 21
24 55 58 05 66 73 99 26 97 17 78</b></span> 78 96 83 14 88 34 89 63 72
21 36 23 09 75 00 76 44 20 45 35 14</b></span> 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
</p>
The product of these numbers is 26 63 78 14 = 1788696.</p>
What is the greatest product of four adjacent numbers in any direction (up, down, left, right, or diagonally) in the 2020 grid?</p>
找出行列连续4个二位数的最大乘积，和problem8有相似的，但复杂一些。</p>
这个跑出的结果也通过了：</p>
#include
int main()
{
int i,j,k,count,tmp1,tmp2,tmp3,tmp4,product=8*2*22*97;
char a[20][40]={0};
char ch[20][59]={
"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"};
for(i=0;i<20;i++)
for(j=0,count=0;j<59;j++)
if(j%3==2)
count++;
else
a[i][j-count]=ch[i][j];
for(i=0;i<17;i++)
for(j=0;j<40;j+=2)
{
for(k=3,tmp1=1;k>=0;k--)
tmp1*=(a[i+k][j]-'0')*10+a[i+k][j+1]-'0';
if(tmp1>product)
product=tmp1;
}
for(i=0;i<20;i++)
for(j=0;j<33;j+=2)
{
for(k=3,tmp2=1;k>=0;k--)
tmp2*=(a[i][j+2*k]-'0')*10+a[i][j+2*k+1]-'0';
if(tmp2>product)
product=tmp2;
}
for(i=0;i<17;i++)
for(j=0;j<33;j+=2)
{
for(k=3,tmp3=1;k>=0;k--)
tmp3*=(a[i+k][j+2*k]-'0')*10+a[i+k][j+2*k+1]-'0';
if(tmp3>product)
product=tmp3;
}
for(i=0;i<17;i++)
for(j=6;j<40;j+=2)
{
for(k=3,tmp4=1;k>=0;k--)
tmp4*=(a[i+k][j-2*k]-'0')*10+a[i+k][j-2*k+1]-'0';
if(tmp4>product)
product=tmp4;
}
printf("the greatest num is %dn",product);
return 0;
}</p>
3.
</p>
Problem 13
</h2>
The sequence of triangle numbers is generated by adding the natural numbers. So the 7th</sup> triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be:</p>
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...</p>
Let us list the factors of the first seven triangle numbers:</p>
1</b>: 1
3</b>: 1,3
6</b>: 1,2,3,6
10</b>: 1,2,5,10
15</b>: 1,3,5,15
21</b>: 1,3,7,21
28</b>: 1,2,4,7,14,28</blockquote>
We can see that 28 is the first triangle number to have over five divisors.</p>
What is the value of the first triangle number to have over five hundred divisors?</p>
</p>
第n个三角形数是前n个自然数的和，求有超过500个因子的最小三角形数：</p>
这个跑了2.3s左右，有更好的方法，欢迎paste上来。</p>
#include
#include
int main()
{
long n,nth,i,j,count;
for(n=7,nth=0;;n++)
{
j=sqrt(nth);
nth=(n+1)*n/2;
for(i=1,count=0;i<=j;i++)
if(nth%i==0)
count+=2;
if(count>500)
break;
}
printf("%ldn",nth);
return 0;
}
</p>
</p>
</p>
</p> </div> </div></div>
分析预测模型在机器学习中的应用之线性回归
回归能做任何事情 Continue reading
android中的线程与进程模型
Published on June 15, 2016
android内存优化之图片加载
Published on June 11, 2016
Find the greatest product of five consecutive digits in the 1000-digit number.</p> 73167176531330624919225119674426574742355349194934 In the 20 08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08 The product of these numbers is 26 What is the greatest product of four adjacent numbers in any direction (up, down, left, right, or diagonally) in the 20 找出行列连续4个二位数的最大乘积，和problem8有相似的，但复杂一些。</p>
这个跑出的结果也通过了：</p>
#include 3. The sequence of triangle numbers is generated by adding the natural numbers. So the 7th</sup> triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be:</p>
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...</p>
Let us list the factors of the first seven triangle numbers:</p> We can see that 28 is the first triangle number to have over five divisors.</p>
What is the value of the first triangle number to have over five hundred divisors?</p>
第n个三角形数是前n个自然数的和，求有超过500个因子的最小三角形数：</p>
这个跑了2.3s左右，有更好的方法，欢迎paste上来。</p>
#include 回归能做任何事情 Continue reading
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450
找出5个连续的数的最大乘积
我正在学C，这是代码。
#include
int main()
{
char ch[1000]=
"73167176531330624919225119674426574742355349194934"
"96983520312774506326239578318016984801869478851843"
"85861560789112949495459501737958331952853208805511"
"12540698747158523863050715693290963295227443043557"
"66896648950445244523161731856403098711121722383113"
"62229893423380308135336276614282806444486645238749"
"30358907296290491560440772390713810515859307960866"
"70172427121883998797908792274921901699720888093776"
"65727333001053367881220235421809751254540594752243"
"52584907711670556013604839586446706324415722155397"
"53697817977846174064955149290862569321978468622482"
"83972241375657056057490261407972968652414535100474"
"82166370484403199890008895243450658541227588666881"
"16427171479924442928230863465674813919123162824586"
"17866458359124566529476545682848912883142607690042"
"24219022671055626321111109370544217506941658960408"
"07198403850962455444362981230987879927244284909188"
"84580156166097919133875499200524063689912560717606"
"05886116467109405077541002256983155200055935729725"
"71636269561882670428252483600823257530420752963450";
int i,temp,product=7*3*1*6*7;
for(i=0;i<996;i++)
{
temp=(ch[i]-'0')*(ch[i+1]-'0')*(ch[i+2]-'0')*(ch[i+3]-'0')*(ch[i+4]-'0');
if(temp>product)
product=temp;
}
printf("%dn",product);
return 0;
}
2.
Problem 11</h2>
20 grid below, four numbers along a diagonal line have been marked in red.</p>
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</b></span> 38 40 67 59 54 70 66 18 38 64 70
67 26 20 68 02 62 12 20 95 63</b></span> 94 39 63 08 40 91 66 49 94 21
24 55 58 05 66 73 99 26 97 17 78</b></span> 78 96 83 14 88 34 89 63 72
21 36 23 09 75 00 76 44 20 45 35 14</b></span> 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
</p>
63 78 14 = 1788696.</p>
20 grid?</p>
int main()
{
int i,j,k,count,tmp1,tmp2,tmp3,tmp4,product=8*2*22*97;
char a[20][40]={0};
char ch[20][59]={
"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"};
for(i=0;i<20;i++)
for(j=0,count=0;j<59;j++)
if(j%3==2)
count++;
else
a[i][j-count]=ch[i][j];
for(i=0;i<17;i++)
for(j=0;j<40;j+=2)
{
for(k=3,tmp1=1;k>=0;k--)
tmp1*=(a[i+k][j]-'0')*10+a[i+k][j+1]-'0';
if(tmp1>product)
product=tmp1;
}
for(i=0;i<20;i++)
for(j=0;j<33;j+=2)
{
for(k=3,tmp2=1;k>=0;k--)
tmp2*=(a[i][j+2*k]-'0')*10+a[i][j+2*k+1]-'0';
if(tmp2>product)
product=tmp2;
}
for(i=0;i<17;i++)
for(j=0;j<33;j+=2)
{
for(k=3,tmp3=1;k>=0;k--)
tmp3*=(a[i+k][j+2*k]-'0')*10+a[i+k][j+2*k+1]-'0';
if(tmp3>product)
product=tmp3;
}
for(i=0;i<17;i++)
for(j=6;j<40;j+=2)
{
for(k=3,tmp4=1;k>=0;k--)
tmp4*=(a[i+k][j-2*k]-'0')*10+a[i+k][j-2*k+1]-'0';
if(tmp4>product)
product=tmp4;
}
printf("the greatest num is %dn",product);
return 0;
}</p>
</p>
Problem 13
</h2>
1</b>: 1
3</b>: 1,3
6</b>: 1,2,3,6
10</b>: 1,2,5,10
15</b>: 1,3,5,15
21</b>: 1,3,7,21
28</b>: 1,2,4,7,14,28</blockquote>
</p>
#include
int main()
{
long n,nth,i,j,count;
for(n=7,nth=0;;n++)
{
j=sqrt(nth);
nth=(n+1)*n/2;
for(i=1,count=0;i<=j;i++)
if(nth%i==0)
count+=2;
if(count>500)
break;
}
printf("%ldn",nth);
return 0;
}
</p>
</p>
</p>
</p> </div> </div></div>分析预测模型在机器学习中的应用之线性回归
android中的线程与进程模型
Published on June 15, 2016
android内存优化之图片加载
Published on June 11, 2016