欧拉计划第６题

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Problem 6

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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.

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平方的和与和的平方之差

前十个自然数的平方的和是

12 + 22 + … + 102 = 385

前十个自然数的和的平方是

(1 + 2 + … + 10)2 = 552 = 3025

因此前十个自然数的平方的和与和的平方之差是 3025 − 385 = 2640。

求前一百个自然数的平方的和与和的平方之差。

 

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代码演示

#include<iostream>
using namespace std;
int main(){
	int sum1=0,sum2=0;
	for(int i=0;i<=100;i++){
		sum1+=i,sum2+=i*i;
	}
	printf("%d\n",sum1*sum1-sum2);
	return 0;
}

 

答案

25164150