There is sequence 1, 12, 123, 1234, ..., 12345678910, ... . Given first N elements of that sequence. You must determine amount of numbers in it that are divisible by 3. Input Input contains N (1<=N<=231 - 1). Output Write answer to the output. Sampl…

一个数能整除3当且仅当各位数之和能整除3. 有了这个规律就好办了, 但是呢,仔细一看, n太大了, 都到 2^31 了.所以简单的模拟肯定不行. 这种貌似像数论的题,一时找不到好办法,就打表! 打表出来是这个样子 1 0    2 1    3 2    4 2    5 3    6 4    7 4    8 5    9 6    10 6 很有规律啊,1,22,3,44…… 如果我们把每三个看成一个(不算1),那么就是每三个元素增加2     于是首先想到是不是 n/3*2 呢? 实验几…

链接: http://vj.acmclub.cn/contest/view.action?cid=168#problem/E 时限:250MS     内存:4096KB     64位IO格式:%I64d & %I64u 提交 状态 练习 SGU 105 问题描述 There is sequence 1, 12, 123, 1234, ..., 12345678910, ... . Given first N elements of that sequence. You must determ…

题目传送门 /* 水题:这题唯一要注意的是要用double,princess可能在一个小时之内被dragon赶上 */ #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> #include <vector> #include <map> using namespace std; ; const int INF = 0x3f3f3f3f;…

分析:很容易知道序列1,2,3, 4,5, 6......与3的关系就是1,2, 0,1, 2,0,......如果是在一个数后面添加一个数就变成了这种序列1, 0, 0, 1, 0, 0, 1, 0, 0.... 代码如下: ======================================================================================= #include<stdio.h> #include<string.h> #inc…

There is sequence 1, 12, 123, 1234, ..., 12345678910, ... . Given first N elements of that sequence. You must determine amount of numbers in it that are divisible by 3. Input Input contains N (1<=N<=231 - 1). Output Write answer to the output. Sampl…

SGU 105-DIV 3 Problem's Link Mean: 定义这样一种数列:1,12,123.. 给出一个n,求这个数列中能被3整除的数的个数. analyse: 这道题可以用分析的方法解决: 对于正整数k,k+1,k+2总有 k+(k+1)+(k+2) =k+k+1+k+2 =3k+3 =3(k+1) 3(k+1)可以被3整除,而且,一个数是否能被3整除表现为它的各位数字之和能否被三整除. 这就意味着对于一个数12345678910111213...k(连写),我们可以从后面开始,…

C. Terse princess time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output «Next please», - the princess called and cast an estimating glance at the next groom. The princess intends to choose the mo…

A. Insomnia cure 哎 只能说英语太差,一眼题我看了三分钟. 题意:给5个数k, l, m, n 和 d,求1~d中能被k, l, m, n 至少一个整除的数的个数. 题解:…… 代码: #include <iostream> using namespace std; int main() { int a, b, c, d, n; cin >> a >> b >> c >> d >> n; ; ; i <= n;…

题目链接:http://www.codeforces.com/problemset/problem/148/A题意:求1到d中有多少个数能被k,l,m,n中的至少一个数整出.C++代码: #include <iostream> using namespace std; int k, l, m, n, d, ans; int main() { cin >> k >>l >> m >> n >> d; ; i <= d; i ++)…