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欧拉计划6-10

By Wu Yudong on January 14, 2017

6、Sum square difference

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.

前十个自然数的平方和是:
12 + 22 + ... + 102 = 385
前十个自然数的和的平方是:
(1 + 2 + ... + 10)2 = 552 = 3025
所以平方和与和的平方的差是3025 - 385 = 2640.

找出前一百个自然数的平方和与和平方的差。

#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;
}

Answer:25164150
Completed on Tue, 2 Apr 2013, 06:57


7、10001st prime

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?

前六个质数是2,3,5,7,11和13,其中第6个是13.

第10001个质数是多少?

#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;
}

Answer:104743

Completed on Thu, 4 Apr 2013, 17:34


8、Largest product in a series

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

找出以下这个1000位的整数中连续5个数字的最大乘积。(例如前五个数字的乘积是7*3*1*6*7=882)

#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;

}

Answer:40824
Completed on Sun, 17 Nov 2013, 11:16


9、Special Pythagorean triplet

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.

一个毕达哥拉斯三元组是一个包含三个自然数的集合,a<b<c,满足条件:
a2 + b2 = c2
例如:32 + 42 = 9 + 16 = 25 = 52.

已知存在并且只存在一个毕达哥拉斯三元组满足条件a + b + c = 1000。

找出该三元组中abc的乘积。

#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;
}

Answer:31875000
Completed on Wed, 24 Jul 2013, 08:53


10、Summation of primes

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.

找出两百万以下所有质数的和。

#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;
}

Answer:142913828922

Completed on Tue, 23 Jul 2013, 17:02

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