Working is not boring, Earning money is boring.
自己做的Project Euler的题解，不见得最优，也不见得优雅，但是答案是对的。基本上可以暴力求解的都是直接暴力求的。
Problem 1:Multiples of 3 and 5
暴力遍历，能整除3或者5的话累加
int problem1()
{
long num = 0;
for (int i = 0; i < 1000; i++)
{
if (i % 3 == 0 || i % 5 == 0)
num += i;
}
return num;
}
Problem 2:Even Fibonacci numbers
把4000000以下的 fib 数列都求出来，再遍历一般累加奇数。
这里其实 fib 数列的奇偶性是有规律的，按照 奇-奇-偶 的形式无限循环，所以也可以用数列的第几项是否能被3整除来判断。
int problem2()
{
long num = 0;
vector<int> fib;
fib.push_back(1);
fib.push_back(2);
while (fib.back() < (4000000))
{
fib.push_back(fib.at(fib.size() - 1) + fib.at(fib.size() - 2));
}
for (auto f : fib)
{
if (f < 4000000) {
cout << f << endl;
if (f % 2 == 0) num += f;
}
}
cout << endl;
return num;
}
Problem 3:Largest prime factor
从小打到大找因数，当找到一个因数之后将原始数字除以这个因数，再继续判断。
long long problem3()
{
vector<long long > nums;
long long num = 600851475143;
long long n = 2;
while (n <= num)
{
if (num % n == 0)
{
nums.push_back(n);
num = num / n;
}
else {
n++;
}
cout << n << endl;
}
cout << endl << endl;
for (auto i : nums)
cout << i << endl;
cout << endl << endl;
return nums.back();
}
Problem 4:Largest palindrome product
暴力遍历，判断是回文数后记录最大值。
int problem4()
{
int num = 0;
int n = 100;
int m = 100;
for (int m = 100; m <= 999; m++)
{
for (int n = 100; n <= 999; n++)
{
stringstream ss;
int nn = n * m;
ss << nn;
string aa = ss.str();
string bb = aa;
reverse(bb.begin(), bb.end());
// cout << aa << " " << bb;
if (aa.compare(bb) == 0)
{
num = MAX(num, n * m);
}
}
cout << m << endl;
}
return num;
}
Problem 5:Smallest multiple
这个没写程序，直接在计算器里乘了一下。。。
Problem 6:Sum square difference
嗯，算数题，算就对了。
int problem6()
{
int num1=0;
int num2 = 0;
for (int i = 1; i <= 100; i++)
{
num1 += i * i;
num2 += i;
}
num2 = num2 * num2;
return num2 - num1;
}
Problem 7:10001st prime
判断素数，前面有一篇文章有详细解释。
int problem7()
{
vector<int> prime;
for (int i = 2; ; i++)
{
int n = 0;
for (int k = 2; k < i; k++)
{
if (i % k == 0) {
n++; break;
}
}
if (n == 0) { prime.push_back(i); cout << i << endl; }
if (prime.size() == 10001) return prime.back();
}
}
Problem 8:Largest product in a series
暴力遍历
int problem8()
{
long long out = 0;
string number = "7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450";
vector<int> numbers;
for (auto i : number)
{
numbers.push_back(i - 48);
}
for (int i = 0; i < numbers.size() - 13; i++)
{
long long num = 1;
for (int k = i; k < i + 13; k++)
{
num = num * numbers.at(k);
}
cout << i <<" "<<out<< endl;
out = MAX(out, num);
}
/*for (int i = 0; i < 13; i++)
{
num = num * numbers.at(i);
out = num;
}
for (int i = 13; i < numbers.size(); i++)
{
num = num * numbers.at(i);
num = num / numbers.at(i - 13);
out = MAX(out, num);
}*/
return out;
}
Problem 9:Special Pythagorean triplet
暴力遍历
void problem9()
{
int a = 1;
int b = 1;
for (a = 1; a < 1000; a++)
{
for (b = 1; b < 1000; b++)
{
int cc = a * a + b * b;
int c = 1000 - a - b;
if (c * c == cc)
cout << "a" << a << "b" << b << "c" << c << "out:"<<a*b*c<<endl;
}
}
}
Problem 10:Summation of primes
同 Problem 7 看之前的文章。
void problem10()
{
vector<int> prime;
long long sum=0;
for (int i = 2; i<2000000; i++)
{
int n = 0;
int ss = sqrt(i);
for (int k = 0; k < prime.size(); k++)
{
if (i % prime.at(k) == 0) {
n++; break;
}
if (prime.at(k) > ss) break;
}
//for (int k = 2; k < i; k++)
//{
// if (i % k == 0) {
// n++; break;
// }
//}
if (n == 0) { prime.push_back(i); sum += i; cout << i << endl;; }
}
cout << sum;
}