欧拉题7
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 10 001st prime number?
第10001个素数
//第n个素数
#include <iostream>
using namespace std;
int IsPrime(int arg)
{
for(int i=2;i<arg;i++)
{
if(arg%i==0)
return 0;
}
return 1;
}
int NstPrime(int n)
{
int index=1,tmp=2;
while(1)
{
if(IsPrime(tmp))
{
if(index==n)
{
return tmp;
}
index++;
}
tmp++;
}
}
int main()
{
int n;
cin>>n;
cout<<NstPrime(n);
return 0;
}欧拉题8
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?
1000个数中连续13个数乘积最大的
//1000个数中连续13个数，最大的乘积
#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <vector>
#define N 1000
using namespace std;
char buffer[2];
vector<int>::iterator it,jt;
int FileRead(char *buffer,vector<int> & number_vec)
{
FILE *f = NULL;
f = fopen("Largest_product_in_a_series.txt","r");
if(NULL == f)
{
cout<<"open failed"<<endl;
return 0;
}
else
cout<<"open success"<<endl;
while(NULL != fgets(buffer,2,f))//buffersize设成2，能一个个数字读
{
//cout<<"buff= "<<buffer<<endl;
number_vec.push_back(atoi(buffer));
}
cout<<"read finished"<<endl;
uint8_t digit[N];
FILE *fp = fopen("Largest_product_in_a_series.txt", "r");
for (int i = 0; i < 1000; i++) {
digit[i] = fgetc(fp) - '0';
//number_vec.push_back(digit[i]);
}
fclose(fp);
for(int i = 0;i<number_vec.size();i++)
{
if(digit[i]!=number_vec[i])
{
cout<<"i failed ="<<i<<endl;
}
}
return 1;
}
int ProductSeries(vector<int> & number_vec,int n)//一开始的思路，得到基础的乘积以后，每次除以前一个乘以后一个，
//会出问题，一方面是0除的问题，另一方面中间出了0后面再乘也还是0，弃用
{
int mul=1,tmp;
for(int i=0;i<n;i++)
{
mul *= number_vec[i];
}
tmp = mul;
cout<<"tmp="<<tmp<<endl;
for(int i=n,j=0;i<number_vec.size();i++,j++)
{
if(0!=number_vec[j])
tmp = tmp/number_vec[j]*number_vec[i];
else
tmp = 0;
//cout<<"tmp="<<tmp<<endl;
if(mul<=tmp)
mul = tmp;
}
return mul;
}
long long int Mul(vector<int> & number_vec)
{
long long int mul=1;
for(it=number_vec.begin();it!=number_vec.end();it++)
{
//cout<<"it="<<*it<<endl;
mul *= (*it);
}
return mul;
}
long long int ProductSeries_imp(vector<int> & number_vec,const int n)
{
FILE *f = NULL;
f = fopen("Largest_product_in_a_series.txt","r");
if(NULL == f)
{
cout<<"open failed"<<endl;
return 0;
}
else
cout<<"open success"<<endl;
int count=0;
while(NULL != fgets(buffer,2,f))//buffersize设成2，能一个个数字读
{
// cout<<"buff= "<<buffer<<endl;
number_vec.push_back(atoi(buffer));
count++;
// cout<<"count="<<count<<endl;
if(n==count)
break;
}
//cout<<"size="<<number_vec.size()<<endl;
long long int mul = Mul(number_vec),tmp;
// cout<<"size="<<number_vec.size()<<endl;
//cout<<mul<<endl;
while(NULL != fgets(buffer,2,f))//buffersize设成2，能一个个数字读
{
tmp = atoi(buffer);
// cout<<"in="<<tmp<<endl;
it = number_vec.begin();
it = number_vec.erase(it);
number_vec.push_back(tmp);
tmp = Mul(number_vec);
if(mul<tmp)
mul = tmp;
}
fclose(f);
return mul;
}
int main()
{
vector<int> number_vec;
long long int mul=1,max=1;
int i,j;
int n=13;
cout<<"max="<<ProductSeries_imp(number_vec,13)<<endl;
number_vec.clear();
if(1!=FileRead(buffer,number_vec))
{
cout<<"failed"<<endl;
return 0;
}
cout<<"size="<<number_vec.size()<<endl;
for(i=0;i<number_vec.size()-n;i++)
{
mul = 1;
for(j = 0;j<n;j++)
{
mul *= number_vec[j+i];//正确答案的位置确实是197后面的数，..
if(i==197)
{
cout<<"197i+j"<<number_vec[j+i]<<endl;
cout<<"mul="<<mul<<endl;//最后一步步算的结果输出，手对比下发现某一步乘法有问题
}
}
if(i==197)
{
cout<<"i=197 "<<mul<<endl;//先测试的197mul算的对不对，发现不对
for(j = 0;j<n;j++)
{
cout<<"i=197 shuzi="<<number_vec[i+j];//然后把书输出来看看读的数对吗，是对的
}
}
if(mul>max)
{
max = mul;
cout<<"i="<<i<<endl;
cout<<"num[i]="<<number_vec[i]<<endl;
cout<<"max="<<max<<endl;
}
// cout<<"i="<<i<<endl;
// cout<<"i+j="<<i+j<<endl;
// cout<<"n="<<n<<endl;
}
cout<<"max2="<<max<<endl;
cout<<sizeof(long long int)<<endl;
}这个题目当时找了一晚上的bug，很蛋疼
查了一晚上欧拉习题8的错误
一开始是实现有问题，没有考虑0的问题，改成一次读n个数进行计算，然后改了以后一直出错，从读文件找，试了其他方法读文件，然后对比，没有问题，又试了全读进去遍历找一遍乘积的方法，再一步步慢慢找到计算得到最大值的位置，再网上找个程序算的位置对比，发现位置有些问题，然后发现网上的这个程序跑出来结果也不对，对着正确答案找位置，自己的程序位置找到的数不对，然后发现网上一个程序的位置时对的，但是数还是错的，就很奇怪了，拿出来计算器手算一遍发现了问题，还是越界的问题，，，，已经考虑过这个问题，把乘积数换成了long int，结果再看一下，发现编译器问题,long int 还是4位的，很尴尬了。。。。查了一晚上逐步定位问题最好找到问题改了。
欧拉题9
Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
a2 + b2 = c2
For example, 32 + 42 = 9 + 16 = 25 = 52.
There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.
满足a+b+c=1000的勾股数，
//满足a+b+c=1000的勾股数
#include <iostream>
using namespace std;
int main()
{
int i,j,tmp;
for(i=3;i<1000;i++)
{
for(j=i;j<1000;j++)
{
tmp = i*i+j*j;
if(tmp>1000*1000) break;
else if((1000-i-j)*(1000-i-j) == tmp)
{
cout<<"i="<<i<<endl;
cout<<"j="<<j<<endl;
cout<<"k="<<1000-i-j<<endl;
}
}
}
cout<<"no apply"<<endl;
} 最直接就是遍历找，然后就想着改进一下的话，b循环从a+1开始，然后a2+b2结果开方，开方出来int再平方对比就知道符合吗，又想到估计开方是泰勒展开做的，可能会耗时间，然后就想着先判断a2+b2没超过1000方，再1000-a-b的平方判断，这样可能会省时间，然后测试发现秒出结果，，