标签:hdu2132
1 2 3 -1
1 3 30
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】
标签:hdu2132
原文地址:http://blog.csdn.net/chang_mu/article/details/40867165