https://www.codechef.com/DEC17/problems/VK18

#include<cstdio>
#include<iostream>
#include<algorithm> using namespace std; #define N 1000001 long long sum[N*],dp[N]; int num[]; void read(int &x)
{
x=; char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) { x=x*+c-''; c=getchar(); }
} int main()
{
int odd,even;
int x,len;
int m=N-<<;
for(int i=;i<=m;++i)
{
odd=even=;
x=i;
len=;
while(x) num[++len]=x%,x/=;
for(int j=;j<=len;++j)
{
if(num[j]&) odd+=num[j];
else even+=num[j];
}
sum[i]=abs(odd-even)+sum[i-];
}
for(int i=;i<N;++i) dp[i]=dp[i-]+(sum[i*]-sum[i])*-(sum[i*]-sum[i*-]);
int T,n;
read(T);
while(T--)
{
read(n);
cout<<dp[n]<<'\n';
}
}

Read problems statements in Mandarin chineseRussian andVietnamese as well.

Chef is so good at programming that he won almost all competitions. With all the prizes, Chef bought a new house. The house looks like a grid of size N (1-indexed) which consists of N × N rooms containing diamonds. For each room, the room number is equal to the sum of the row number and the column number.

The number of diamonds present in each room is equal to the absolute difference between the sum of even digits and sum of odd digits in its room number. For example, if the room number is 3216, then the number of diamonds present in that room will be |(2+6)-(3+1)| = 4.

You are given the number N. You have to print the total number of diamonds present in Chef's house.

Input

  • The first line of the input contains a single integer T denoting the number of test cases. The description of T test cases follows.
  • The only line of each test case contains a single integer N.

Output

For each test case, print the answer on a separate line.

Constraints

  • 1 ≤ T ≤ 105
  • 1 ≤ N ≤ 106

Subtasks

Subtask #1 (15 points):

  • 1 ≤ T ≤ 10
  • 1 ≤ N ≤ 1000

Subtask #2 (15 points):

  • 1 ≤ T ≤ 10
  • 1 ≤ N ≤ 106

Subtask #3 (70 points): original constraints

Example

Input:

3
1
2
3 Output: 2
12
36

CF&&CC百套计划2 CodeChef December Challenge 2017 Total Diamonds的更多相关文章

  1. CF&&CC百套计划2 CodeChef December Challenge 2017 Chef And Easy Xor Queries

    https://www.codechef.com/DEC17/problems/CHEFEXQ 题意: 位置i的数改为k 询问区间[1,i]内有多少个前缀的异或和为k 分块 sum[i][j] 表示第 ...

  2. CF&&CC百套计划2 CodeChef December Challenge 2017 Chef and Hamming Distance of arrays

    https://www.codechef.com/DEC17/problems/CHEFHAM #include<cstdio> #include<cstring> #incl ...

  3. CF&&CC百套计划2 CodeChef December Challenge 2017 Penalty Shoot-out

    https://www.codechef.com/DEC17/problems/CPLAY #include<cstdio> #include<algorithm> using ...

  4. CF&&CC百套计划2 CodeChef December Challenge 2017 Chef And his Cake

    https://www.codechef.com/DEC17/problems/GIT01 #include<cstdio> #include<algorithm> using ...

  5. CF&&CC百套计划4 Codeforces Round #276 (Div. 1) A. Bits

    http://codeforces.com/contest/484/problem/A 题意: 询问[a,b]中二进制位1最多且最小的数 贪心,假设开始每一位都是1 从高位i开始枚举, 如果当前数&g ...

  6. CF&&CC百套计划4 Codeforces Round #276 (Div. 1) E. Sign on Fence

    http://codeforces.com/contest/484/problem/E 题意: 给出n个数,查询最大的在区间[l,r]内,长为w的子区间的最小值 第i棵线段树表示>=i的数 维护 ...

  7. CF&&CC百套计划1 Codeforces Round #449 C. Willem, Chtholly and Seniorious (Old Driver Tree)

    http://codeforces.com/problemset/problem/896/C 题意: 对于一个随机序列,执行以下操作: 区间赋值 区间加 区间求第k小 区间求k次幂的和 对于随机序列, ...

  8. CF&&CC百套计划3 Codeforces Round #204 (Div. 1) D. Jeff and Removing Periods

    http://codeforces.com/problemset/problem/351/D 题意: n个数的一个序列,m个操作 给出操作区间[l,r], 首先可以删除下标为等差数列且数值相等的一些数 ...

  9. CF&&CC百套计划1 Codeforces Round #449 A. Nephren gives a riddle

    http://codeforces.com/contest/896/problem/A 第i个字符串嵌套第i-1个字符串 求第n个字符串的第k个字母 dfs #include<map> # ...

随机推荐

  1. 生命周期事件和 Global.asax 文件

    文章:IIS 5.0 和 6.0 的 ASP.NET 应用程序生命周期概述 该文章有介绍Application_Start方法

  2. Internet History, Technology and Security (Week⑨)

    Week ⑨ We are now on the second to last week of the class and finishing up our look at Internet Secu ...

  3. ajax 数据请求(一)同域

    参考:http://www.css88.com/jqapi-1.9/jQuery.ajax/ http://www.cnblogs.com/haitao-fan/p/3908973.html 1.常用 ...

  4. PHP 常用函数总结(一)

    重点: 看函数的 ‘参数’ 和 ‘返回值’ 文档:http://www.w3school.com.cn/php/php_ref_array.asp http://www.w3school.com.cn ...

  5. 高性能页面加载技术--BigPipe设计原理及Java简单实现

    1.技术背景 动态web网站的历史可以追溯到万维网初期,相比于静态网站,动态网站提供了强大的可交互功能.经过几十年的发展,动态网站在互动性和页面显示效果上有了很大的提升,但是对于网站动态网站的整体页面 ...

  6. delphi locate函数的使用

    loc1:= qry1.FieldbyName('SPBM').AsString;      //商品编码 loc2:= qry1.FieldbyName('XH').AsString;       ...

  7. Ubuntu 12.04下LAMP环境搭建实录

    方法一: 用tasksel可以方便安装dns server, lamp, kubuntu desktop, ubuntu desktop, xubuntu之类的软件包.这个软件在ubuntu serv ...

  8. Python画统计图

    https://blog.csdn.net/jenyzhang/article/details/52046372

  9. 【刷题】洛谷 P3950 部落冲突

    题目背景 在一个叫做Travian的世界里,生活着各个大大小小的部落.其中最为强大的是罗马.高卢和日耳曼.他们之间为了争夺资源和土地,进行了无数次的战斗.期间诞生了众多家喻户晓的英雄人物,也留下了许多 ...

  10. DB2 Vs MySQL系列 | MySQL与DB2的数据类型对比

    随着MySQL数据库的应用越来越广泛,DB2向MySQL数据库的迁移需求也越来越多.进行数据库之间迁移的时候,首先遇到的并且也是最基本最重要的就是两种数据库数据类型之间的转换. 相关阅读: 从商用到开 ...