An easy problem

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import java.util.Scanner;
import java.math.BigInteger;

public class Main {
    static final int maxn = 100005;
    static BigInteger[] arr = new BigInteger[maxn];

    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in);
        arr[0] = BigInteger.ZERO;
        BigInteger tmp;
        for(int i = 1; i <= 100000; ++i) {
            tmp = BigInteger.valueOf(i);
            if(i % 3 == 0)
                arr[i] = arr[i-1].add(tmp.multiply(tmp).multiply(tmp));
            else arr[i] = arr[i-1].add(tmp);
        }
        int n;
        while(true) {
            n = cin.nextInt();
            if(n < 0) break;
            System.out.println(arr[n]);
        }
    }
}

#include <stdio.h>

#define maxn 100005
typedef __int64 LL;

LL arr[maxn];

int main() {
    int n, i;
    for(i = 1; i <= 100000; ++i)
        if(i % 3) arr[i] = arr[i-1] + i;
        else arr[i] = arr[i-1] + (LL)i * i * i;
    while(scanf("%d", &n), n >= 0)
        printf("%I64d\n", arr[n]);
}

HDU2132 An easy problem 【Java】

原文：http://blog.csdn.net/chang_mu/article/details/40867165