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  • HDOJ(HDU) 2132 An easy problem

    Problem Description
    We once did a lot of recursional problem . I think some of them is easy for you and some if hard for you.
    Now there is a very easy problem . I think you can AC it.
    We can define sum(n) as follow:
    if i can be divided exactly by 3 sum(i) = sum(i-1) + i*i*i;else sum(i) = sum(i-1) + i;
    Is it very easy ? Please begin to program to AC it..-_-

    Input
    The input file contains multilple cases.
    Every cases contain only ont line, every line contains a integer n (n<=100000).
    when n is a negative indicate the end of file.

    Output
    output the result sum(n).

    Sample Input
    1
    2
    3
    -1

    Sample Output
    1
    3
    30

    水题。。注意范围。!!!java用long型可以AC,只是注意中间计算结果也有可能溢出int型范围,也要转换为long才行。
    还有,注意判断条件退出不是输入-1,而是输入小于0的数就是退出了。

    import java.util.Scanner;
    
    public class Main{
        static long db[] = new long[100001];
        public static void main(String[] args) {
            dabiao();
            Scanner sc = new Scanner(System.in);
            while(sc.hasNext()){
                int n =sc.nextInt();
                if(n<0){
                    return;
                }
                System.out.println(db[n]);
            }
        }
        private static void dabiao() {
            db[1]=1;
            db[2]=3;
            for(int i=3;i<db.length;i++){
                if(i%3==0){
                    db[i]=db[i-1]+i*(long)i*i;
                    //这里的i*i要强转成long,long*int还是long,否则i*i*i会超int范围
                }else{
                    db[i]=db[i-1]+i;
                }
            }
    
        }
    
    }
    
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  • 原文地址:https://www.cnblogs.com/webmen/p/5739233.html
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