欧拉计划006

Sum square difference

The sum of the squares of the first ten natural numbers is,

\[1^2+2^2+…+10^2=385 \]

The square of the sum of the first ten natural numbers is,

\[(1+2+…+10)^2=55^2=3025 \]

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.

平方的和与和的平方之差

前十个自然数的平方的和是

\[1^2+2^2+…+10^2=385 \]

前十个自然数的和的平方是

\[(1+2+…+10)^2=55^2=3025 \]

因此，前十个数的平方的和与和的平方之差是$$3025−385=2640$$。

求前一百个数的平方的和与和的平方之差。

这道题目很简单，只要分别求解一下相减就好了。求解出来答案为25164150。

#include<iostream>
using namespace std;

int main(){
    int sum1=0;
    for(int i=1;i<=100;i++){
        sum1 = sum1 + i*i;
    }
    int sum2=0;
    for (int i=1;i<=100;i++){
        sum2 = sum2+i;
    }
    sum2 = sum2*sum2;
    cout<<"两数之差为"<<sum2-sum1<<endl;
    return 0;
}