欧拉#8 的解决方案
#include <stdio.h>
int main(void)
{
char *num = "73167176531330624919225119674426574742355349194934"
"96983520312774506326239578318016984801869478851843"
"85861560789112949495459501737958331952853208805511"
"12540698747158523863050715693290963295227443043557"
"66896648950445244523161731856403098711121722383113"
"62229893423380308135336276614282806444486645238749"
"30358907296290491560440772390713810515859307960866"
"70172427121883998797908792274921901699720888093776"
"65727333001053367881220235421809751254540594752243"
"52584907711670556013604839586446706324415722155397"
"53697817977846174064955149290862569321978468622482"
"83972241375657056057490261407972968652414535100474"
"82166370484403199890008895243450658541227588666881"
"16427171479924442928230863465674813919123162824586"
"17866458359124566529476545682848912883142607690042"
"24219022671055626321111109370544217506941658960408"
"07198403850962455444362981230987879927244284909188"
"84580156166097919133875499200524063689912560717606"
"05886116467109405077541002256983155200055935729725"
"71636269561882670428252483600823257530420752963450";
int i, tmp=1, product = 0;
for(i = 0; num[i] != NULL; i++)
{
tmp *= (num[i] - '0');
if((i+1) % 5 == 0)
{
if(tmp > product)
product = tmp;
tmp = 1;
}
}
printf("Largest product ------> %i\n", product);
return 0;
}
目的是找出这1000位数字中五个连续数字的最大乘积。答案是 40824,但我的解决方案产生 31752。关于我哪里出错了,有什么想法吗?
#include <stdio.h>
int main(void)
{
char *num = "73167176531330624919225119674426574742355349194934"
"96983520312774506326239578318016984801869478851843"
"85861560789112949495459501737958331952853208805511"
"12540698747158523863050715693290963295227443043557"
"66896648950445244523161731856403098711121722383113"
"62229893423380308135336276614282806444486645238749"
"30358907296290491560440772390713810515859307960866"
"70172427121883998797908792274921901699720888093776"
"65727333001053367881220235421809751254540594752243"
"52584907711670556013604839586446706324415722155397"
"53697817977846174064955149290862569321978468622482"
"83972241375657056057490261407972968652414535100474"
"82166370484403199890008895243450658541227588666881"
"16427171479924442928230863465674813919123162824586"
"17866458359124566529476545682848912883142607690042"
"24219022671055626321111109370544217506941658960408"
"07198403850962455444362981230987879927244284909188"
"84580156166097919133875499200524063689912560717606"
"05886116467109405077541002256983155200055935729725"
"71636269561882670428252483600823257530420752963450";
int i, tmp=1, product = 0;
for(i = 0; num[i] != NULL; i++)
{
tmp *= (num[i] - '0');
if((i+1) % 5 == 0)
{
if(tmp > product)
product = tmp;
tmp = 1;
}
}
printf("Largest product ------> %i\n", product);
return 0;
}
The object is to find the largest product of five consecutive numbers in this 1,000 digit number. The answer is 40824, however my solution produces 31752. Any ideas as to where I have gone wrong?
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您只能找到每 5 组 5 个连续数字的最高乘积,而不是每组 5 个连续数字。您只查看了 20% 的可能组合。
例如,在第一行数据上,您看到的第一个序列是“96983”。第二个是“52031”。
您错过了“69835”、“98352”、“83520”和“35203”。
You are only finding the highest product of each 5th set of 5 consecutive digits, not every set of 5 consecutive digits. You are only looking at 20% of the possible combinations.
e.g. on the first line of data, the first sequence you look at is "96983". The second is "52031".
You miss "69835", "98352", "83520" and "35203".
您应该通过删除存在的“\n”将 1000 位数字转换为一个连续的数字串,而不是像现在这样使用单独的行。然后您将看到一个 5 位数字的窗口,在该窗口中的数字您将找到产品,如果产品是迄今为止最大的产品,您可以保存它,然后将窗口向下滑动一位数字并重复直到结束。这应该会给你留下答案。
You should turn the 1000 digit number into one consecutive string of digits by removing the "\n" present instead of having separate lines like you do now. Then you will have a 5 digit window and the numbers within this window you will find the product and if the product is the largest thus far you save it and then slide the window down one digit and repeat until the end. This should leave you with your answer.