Educational Codeforces Round 23C
超级坑的水题!!!想了两天没一点思路,看了题解第一段话就做出来了
刚开始一直在想找到通项就是例如an*10^n+...+a0*10^0-an-...-a0>=s,然后从这个里面找到规律,结果走进死胡同了
Let's prove that if x is really big, then x + 1 is really big too.
证明很容易,我就不证了,直接上代码,二分答案就行了
还有一点要注意,因为是10E18范围,所以平时循环50次是不够用的,加到了70次就ac了,也可以用l<r-1这个来判断
#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<cassert>
#include<iomanip>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define pi acos(-1)
#define ll long long
#define mod 1000000007
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#pragma comment(linker, "/STACK:1024000000,1024000000") using namespace std; const double g=10.0,eps=1e-;
const int N=+,maxn=,inf=; ll n,s;
bool ok(ll x)
{
ll t=x,p=x;
while(p){
t-=(p%);
p/=;
}
return t>=s;
}
int main()
{
ios::sync_with_stdio(false);
cin.tie();
cin>>n>>s;
ll l=,r=n+;
for(int i=;i<;i++)
{
ll m=(l+r)/;
if(ok(m))r=m;
else l=m;
}
cout<<n-l<<endl;
return ;
}
Educational Codeforces Round 23C的更多相关文章
- [Educational Codeforces Round 16]E. Generate a String
[Educational Codeforces Round 16]E. Generate a String 试题描述 zscoder wants to generate an input file f ...
- [Educational Codeforces Round 16]D. Two Arithmetic Progressions
[Educational Codeforces Round 16]D. Two Arithmetic Progressions 试题描述 You are given two arithmetic pr ...
- [Educational Codeforces Round 16]C. Magic Odd Square
[Educational Codeforces Round 16]C. Magic Odd Square 试题描述 Find an n × n matrix with different number ...
- [Educational Codeforces Round 16]B. Optimal Point on a Line
[Educational Codeforces Round 16]B. Optimal Point on a Line 试题描述 You are given n points on a line wi ...
- [Educational Codeforces Round 16]A. King Moves
[Educational Codeforces Round 16]A. King Moves 试题描述 The only king stands on the standard chess board ...
- Educational Codeforces Round 6 C. Pearls in a Row
Educational Codeforces Round 6 C. Pearls in a Row 题意:一个3e5范围的序列:要你分成最多数量的子序列,其中子序列必须是只有两个数相同, 其余的数只能 ...
- Educational Codeforces Round 9
Educational Codeforces Round 9 Longest Subsequence 题目描述:给出一个序列,从中抽出若干个数,使它们的公倍数小于等于\(m\),问最多能抽出多少个数, ...
- Educational Codeforces Round 37
Educational Codeforces Round 37 这场有点炸,题目比较水,但只做了3题QAQ.还是实力不够啊! 写下题解算了--(写的比较粗糙,细节或者bug可以私聊2333) A. W ...
- Educational Codeforces Round 60 (Rated for Div. 2) - C. Magic Ship
Problem Educational Codeforces Round 60 (Rated for Div. 2) - C. Magic Ship Time Limit: 2000 mSec P ...
随机推荐
- oracle的认证方式
使用as sysdba是使用操作系统验证方式,不需要输入密码
- ZOJ 3537 Cake(凸包判定+区间DP)
Cake Time Limit: 1 Second Memory Limit: 32768 KB You want to hold a party. Here's a polygon-shaped c ...
- C++ Design Pattern: What is a Design Pattern?
Q: What is a Design Pattern? A: Design Patterns represent solutions to problems what arise when deve ...
- Day06 DOM4J&schema介绍&xPath
day06总结 今日内容 XML解析之JAXP( SAX ) DOM4J Schema 三.XML解析器介绍 操作XML文档概述 1 如何操作XML文档 XML文档也是数据的一种,对数据的 ...
- Struts2表单数据接收方式
版权声明:本文为博主原创文章,未经博主同意不得转载. https://blog.csdn.net/sunshoupo211/article/details/30249239 1.将Action类作 ...
- 205-react SyntheticEvent 事件
参看地址:https://reactjs.org/docs/events.html
- nodejs Async详解之二:工具类
Async中提供了几个工具类,给我们提供一些小便利: memoize unmemoize log dir noConflict 1. memoize(fn, [hasher]) 有一些方法比较耗时,且 ...
- 翻译:Addressing tiles: same tile bounds with different indexes
原文链接:http://www.maptiler.org/google-maps-coordinates-tile-bounds-projection/ Addressing tiles: same ...
- makefile中ifeq与ifneq dev/null和dev/zero简介 dd命令
ifeq语法是ifeq "<arg1>;" "<arg2>;" ,功能是比较参数“arg1”和“arg2”的值是否相同,相同时为1 i ...
- SpringData_JpaRepository接口
该接口提供了JPA的相关功能 List<T> findAll(); //查找所有实体 List<T> findAll(Sort sort); //排序.查找所有实体 List& ...