problem 8 :Largest product in a series
The four adjacent digits in the 1000-digit number that have the greatest product are 9 × 9 × 8 × 9 = 5832.
73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450
Find the thirteen adjacent digits in the 1000-digit number that have the greatest product. What is the value of this product?
思路 :
题目的意思是这是一个1000位的数,其中相邻的四位数相乘所得最大的数为5832,现在让我们求相邻的十三位数相乘所得最大的数是多少?
首先我们从题目的那个例子里思考入手,应该是循环相乘比较,并求出最大得数。我们思考问题可以由易到难,我们假设有一个十位数的数,随便写一个十位数的数--------5265659989,接下来我们设计一下程序的算法,画一张图解:
这是一个十位数的数,我们求相邻四位数相乘所得的最大数,相当于我们将其分割为6个块(相当于1到10-4+1),设置一个变量max,首先将其赋值为0,循环变量 i 从1到10-4+1,譬如第一次算第一位乘到第四位(5x2x6x5)的值赋给变量value, 每算出一次value,将其与max比较,若value大于max,就将value的值赋值给max,否则就继续执行下一次,这样我们最后得到的max就是最大值输出。完了谢谢!
代码 :
%Problem 8
tic
w = '7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450';
%%just copy the numbers to here
max = 0; %initial value
value = 1; %initial value
for i = 1:988
n1 = str2double(w(i));
n2 = str2double(w(i+1));
n3 = str2double(w(i+2));
n4 = str2double(w(i+3));
n5 = str2double(w(i+4));
n6 = str2double(w(i+5));
n7 = str2double(w(i+6));
n8 = str2double(w(i+7));
n9 = str2double(w(i+8));
n10 = str2double(w(i+9));
n11= str2double(w(i+10));
n12 = str2double(w(i+11));
n13 = str2double(w(i+12));
value = n1*n2*n3*n4*n5*n6*n7*n8*n9*n10*n11*n12*n13;
if value > max
max = value; %update the value of max
end
end
disp(max)
toc
结果 : 2.351462400000000e+10
关于结果的输出这里你需要在matlab的菜单栏里找到Preferences--->Command Windows,找到右边的Text display,将Numeric format中的short改为long e。
小结 :
希望大家多多交流,互相学习!