Project Euler -> problem 6
2013-08-22 15:33
330 查看
6. 前十个自然数的平方和是:
1^2 + 2^2 + ... + 10^2 = 385
前十个自然数的和的平方是:
(1 + 2 + ... + 10)^2 = 55^2 = 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。
int main(void)
{
long
i,j=0,k=0;
for(i=1;i<=100;i++){
j =
i*i+j;
}
for(i=0;i<=100;i++){
k+=i;
}
printf("The result
is:%ld",k*k-j);
return
0;
}
1^2 + 2^2 + ... + 10^2 = 385
前十个自然数的和的平方是:
(1 + 2 + ... + 10)^2 = 55^2 = 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。
int main(void)
{
long
i,j=0,k=0;
for(i=1;i<=100;i++){
j =
i*i+j;
}
for(i=0;i<=100;i++){
k+=i;
}
printf("The result
is:%ld",k*k-j);
return
0;
}
Answer: | 25164150 |
相关文章推荐
- Project Euler -> problem 2
- Project Euler -> problem 3
- Project Euler -> problem 4
- Project Euler -> problem 5
- Project Euler -> problem 7
- Project Euler -> problem 8
- Project Euler -> problem 9
- Project Euler -> problem 12
- Project Euler -> problem 1
- Project Euler -> prob 4000 lem 11
- project euler Problem 18 & 36
- projecteuler---->problem=33----Digit canceling fractions
- Project Euler Problem 30
- ZOJ 3686 A Simple Tree Problem(将对树的操作转化成区间=>线段树)
-  |"|&|<|>等html字符转义
- The7th Zhejiang Provincial Collegiate Programming Contest->Problem A:A - Who is Older?
- Test11_6&nbsp;指针用p-&gt;X和(*p).X等价
- JSP 之 JSTL_04:<c:forTokens>
- Linux-world-2012-January->16(mini2440 uboot201103 系统移植)->3(machine ID problem)
- project euler Largest prime factor&Largest palindrome product