# 工学1号馆

home

## 欧拉计划6-10

By Wu Yudong on January 14, 2017

The sum of the squares of the first ten natural numbers is,
12 + 22 + ... + 102 = 385
The square of the sum of the first ten natural numbers is,
(1 + 2 + ... + 10)2 = 552 = 3025
Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 − 385 = 2640.

Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.

12 + 22 + ... + 102 = 385

(1 + 2 + ... + 10)2 = 552 = 3025

#include <stdio.h>
#include <string.h>
#include <ctype.h>
#include <math.h>

#define N 100

int powplus(int n, int k)
{
int s = 1;
while (k--) {
s *= n;
}
return s;
}

int sum1(int n)
{
return powplus((n + 1) * n / 2, 2);
}

int sum2(int n)
{
return (n * (n + 1) * (2 * n + 1)) / 6;
}

void solve()
{
printf("%d\n", sum1(N));
printf("%d\n", sum2(N));
printf("%d\n", sum1(N) - sum2(N));
}

int main()
{
solve();
return 0;
}

Completed on Tue, 2 Apr 2013, 06:57

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?

#include <stdio.h>
#include <string.h>
#include <ctype.h>
#include <math.h>

int prim(int n)
{
int i;
for (i = 2; i * i <= n; i++) {
if (n % i == 0)
return 0;
}
return 1;
}

void solve(int n)
{
int i = 2;
int count = 0;
while (1) {
if (prim(i)) {
count++;
if (count == n)
break;
}
i++;
}
printf("%d\n", i);
}

int main()
{
int n = 10001;
solve(n);
return 0;
}

Completed on Thu, 4 Apr 2013, 17:34

Find the greatest product of five consecutive digits in the 1000-digit number.

73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450

#include <stdio.h>
#include <stdlib.h>

int main()
{
FILE *fp;
char *buffer;
int i = 0, j = 0;

fp = fopen("E://file.txt", "r");

char c;

while ((c = fgetc(fp))) {
if (c == EOF)
break;
else if (c != '\n')
i++;
}
buffer = (char *)malloc(i * sizeof(char));
rewind(fp);
while ((c = fgetc(fp))) {
if (c == EOF)
break;
else if (c != '\n') {
*(buffer + j) = c;
j++;
}
}
findmax(buffer, i);
}

int findmax(char *buffer, int i)
{
int j = 0, max = 0;

for (j = 0; j < i - 4; j++) {
if (max < ((buffer[j] - '0') * (buffer[j + 1] - '0') * (buffer[j + 2] - '0') * (buffer[j + 3] - '0') * (buffer[j + 4] - '0')))
max = ((buffer[j] - '0') * (buffer[j + 1] - '0') * (buffer[j + 2] - '0') * (buffer[j + 3] - '0') * (buffer[j + 4] - '0'));

}
printf("\n%d", max);
return 0;

}

Completed on Sun, 17 Nov 2013, 11:16

A 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.

a2 + b2 = c2

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<ctype.h>
#include<stdlib.h>
#include<stdbool.h>

void show()
{
int a, b, c;
for (a = 1; a < 333; a++) {
for (c = 333; c < 500; c++) {
b = 1000 - a - c;
if (a * a + b * b == c * c) {
printf("%d\n", a * b * c);
return;
}
}
}
}

int main()
{
show();
return 0;
}

Completed on Wed, 24 Jul 2013, 08:53

The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.

Find the sum of all the primes below two million.

10以下的质数的和是2 + 3 + 5 + 7 = 17.

#include<stdio.h>
#include<math.h>
#include<stdbool.h>

#define N 2000000

bool prim(int n)
{
int i;
for (i = 2; i * i <= n; i++) {
if (n % i == 0)
return false;
}
return true;
}

int main()
{
int i;
long long sum = 2;
for (i = 3; i <= N; i = i + 2) {
if (prim(i)) {
sum += i;
}
}
printf("%lld\n", sum);

return 0;
}