求大身做做这个oj题目

#include<iostream>
using namespace std;
int oddSum(int n) {
    return n * n * (2 * n * n - 1);
}
int evenSum(int n) { 
    return 2 * 2 * n * (n + 1) * (2*n + 1) / 6;
}
int main() {
    int N, m, n;
    cin >> N;
    while (N--) {
        cin >> m >> n;
        if (m > n) swap(m, n);
        // sum(x^2) | m n
        //  1^2 + ... + n^2 = n*(n+1)*(2n+1)/6
        // (2*1)^2 + ... + (2*n)^2 = 4 *  ( 1^2 + ... + n^2) = 4 * n*(n+1)*(2n+1)/6
        cout << evenSum(n) - evenSum(m) << endl;
        // 奇数sum(x^3) | m n
        //  (2*1-1)^3 + ... + (2*n-1)^3 = n^2 * (2*n^2 -1)
        cout << oddSum(n) - oddSum(m) << endl;
    }
}

#include <bits/stdc++.h>

using namespace std;

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t;
    int m, n;
    cin >> t;
    while(t--){
        cin >> n >> m;
        int ans = 0;
        if(n > m) swap(n, m);
        for(int i=n;i<=m;i++){
            if(i & 1) ans += i * i * i;
            else ans += i * i;
        }
        cout << ans << '\n';
    }
    return 0;
}

下面作为给其他读者参考的答案吧，其实题主采纳的是下面的代码，之前回答考虑少了，然后输出的不是正确的，现在才对

#include<iostream>
using namespace std;
int oddSum(int n) {
    return n * n * (2 * n * n - 1);
}
int evenSum(int n) { 
    return 2 * 2 * n * (n + 1) * (2*n + 1) / 6;
}
int main() {
    int N, m, n;
    int minOdd, maxOdd, minEven, maxEven;
    cin >> N;
    while (N--) {
        cin >> m >> n;
        if (m > n) swap(m, n);

        // sum(x^2) | m n
        //  1^2 + ... + n^2 = n*(n+1)*(2n+1)/6
        // (2*1)^2 + ... + (2*n)^2 = 4 *  ( 1^2 + ... + n^2) = 4 * n*(n+1)*(2n+1)/6
        minEven = m-1;
        while (minEven>0&&minEven % 2 == 1) {
            minEven -= 1;
        }
        if (minEven < 0) minEven = 0;
        maxEven = n;
        while (maxEven % 2 == 1) {
            maxEven -= 1;
        }
        if (maxEven < 0) maxEven = 0;
        // 转换成序号
        cout << - evenSum(minEven/2) + evenSum(maxEven/2) << " ";
        
        // 奇数sum(x^3) | m n
        //  (2*1-1)^3 + ... + (2*n-1)^3 = n^2 * (2*n^2 -1)
        minOdd = m-1;
        while (minOdd>0&&minOdd % 2 == 0) {
            minOdd -= 1;
        }
        if (minOdd < 0) minOdd = 0;
        maxOdd = n;
        while (maxOdd % 2 == 0) {
            maxOdd -= 1;
        }
        if (maxOdd < 0) maxOdd = 0;
        // 转换成序号
        cout <<  - oddSum((minOdd+1)/2 ) + oddSum((maxOdd + 1) / 2) << endl;
    }
}

#include <stdio.h>
#include <math.h>
int main()
{

    int a = 0,b = 0,i = 0,count,tmp,j=0;
    int arr[64];
    printf("输入几组数据？:\n");
    scanf("%d",&count);
    tmp = count;
    printf("输入每组的两个整数：\n");
    while(tmp--)
    {   
        scanf("%d %d",&arr[i],&arr[i+1]);
        i += 2;
    }   

    for(j-0;j<count*2;j++)
    {   
        if(arr[j]%2)
        {
            a += pow(arr[j],3);
        }
        else
        {
            b +=pow(arr[j],2);
        }
    }   

    printf("所有偶数的平方和为:%d\n所有奇数的三次方和为:%d\n",b,a);
    return 0;
}