zoukankan      html  css  js  c++  java
  • hdoj 1977 Consecutive sum II

    Consecutive sum II

    Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 2523    Accepted Submission(s): 1219


    Problem Description
    Consecutive sum come again. Are you ready? Go ~~
    1    = 0 + 1
    2+3+4    = 1 + 8
    5+6+7+8+9  = 8 + 27

    You can see the consecutive sum can be representing like that. The nth line will have 2*n+1 consecutive numbers on the left, the first number on the right equal with the second number in last line, and the sum of left numbers equal with two number’s sum on the right.
    Your task is that tell me the right numbers in the nth line.
     
    Input
    The first integer is T, and T lines will follow.
    Each line will contain an integer N (0 <= N <= 2100000).
     
    Output
    For each case, output the right numbers in the Nth line.
    All answer in the range of signed 64-bits integer.
     
    Sample Input
    3
    0
    1
    2
     
    Sample Output
    0 1
    1 8
    8 27
     
    找规律
    #include<stdio.h>
    #include<string.h>
    int main()
    {
    	long long m,n;
    	scanf("%lld",&n);
    	while(n--)
    	{
    		scanf("%lld",&m);
    		printf("%lld %lld
    ",m*m*m,(m+1)*(m+1)*(m+1));
    	}
    	return 0;
    } 
    

      

  • 相关阅读:
    移动Web开发规范概述
    hibernate 多对多
    hibernate 1 对1
    hibernate 双向1对多
    Hibernate 单项多对1
    Hibernate Session 4种对象状态
    Hibernate Session缓存
    Hibernaate 详解
    Hibernate学习 (一)
    Struts拦截器Interceptor
  • 原文地址:https://www.cnblogs.com/tonghao/p/4579378.html
Copyright © 2011-2022 走看看