欧拉计划(6)Sum square difference
2015-05-09 16:22
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【题目】
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.
【翻译】
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。
【思路】直接套公式,简单题。
【代码】
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.
【翻译】
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。
【思路】直接套公式,简单题。
【代码】
void test6() { int n=100; int sum1=(n*n)*((n+1)*(n+1))/4; int sum2=(n*(n+1)*((2*n)+1))/6; cout<<sum1-sum2<<endl; }【答案】本题答案为25164150 。
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