hdu 2451 Simple Addition Expression

Problem Description
A luxury yacht with 100 passengers on board is sailing on the sea in the twilight. The yacht is ablaze with lights and there comes out laughers and singing from the hall where an evening party is in full swing. People are singing, dancing and enjoying themselves.

The yacht is equipped with the most advanced navigation and driving system which can all be manipulated by a computer. When the captain notices that there is only gentle breeze and the sea waves are not high, he starts the autopilot. The yacht sails forward smoothly, ploughs the waves. When it’s completely dark, the passengers start to feel a little funny for sudden forward rushes or sudden decelerations or slight swings. The captain immediately walks to the driving platform and switches the autopilot to human manipulation. The yacht returns back to normal and the party restarts. Laughers come back, too.

The captain summons the engineer on board to do a thorough check of the navigation system. It turns out that only the computer is out of order, but the exact failure is still unclear. There is a computer scientist among the passengers who is also invited to the cab to give a hand. He first inputs several groups of data to test the computer. When he inputs 1+2+3, the computer outputs 6, which is exactly right. But when he inputs 4+5+6, the computer outputs 5, which is wrong. Then he inputs 12+13+14, and gets 39, another right answer, while he inputs 14+15+16, and gets 35, another wrong answer. After the test, the computer scientist says smilingly: “the failure is clear now. The computer's adder can not carry." After excluding the failure, the captain restarts the autopilot and the yacht returns back to normal, sailing smoothly on the sea.

The captain and the engineer invite the computer scientist to sit down and have a talk. The computer scientist tells a story as following:

A former mathematician defined a kind of simple addition expression. 

If there is an expression (i) + (i+1) + (i+2), i>=0, when carried out additive operations, no position has a carry, it is called simple addition expression.

For instance, when i equals 0, 0+1+2 is a simple addition expression, meanwhile when i equals 11, 11+12+13 is a simple addition expression, too. Because of that no position has a carry.

However, when i equals 3, 3+4+5 is not a simple addition expression, that is because 3+4+5 equals 12, there is a carried number from unit digit to tens digit. In the same way, when i equals 13, 13+14+15 is not a simple addition expression, either. However, when i equals 112, 112+113+114 is a simple addition expression. Because 112+113+114 equals 339, there is no carry in the process of adding.

when the students have got the definition of simple addition expression, the mathematician puts forward a new question: for a positive integer n, how many simple addition expressions exist when i<n. In addition, i is the first number of a simple addition expression.

when the value of n is large enough, the problem needs to be solved by means of computer.

 
Input
There are several test cases, each case takes up a line, there is an integer n (n<10^10).

 
Output
Output the number of all simple addition expressions when i<n.

 
Sample Input
1
2
3
4
10
11
 
Sample Output
1
2
3
3
3
4
 
Source
 
Recommend
gaojie
 

题意:给出任意N,求从0~N-1这N个数中找出 N+(N+1)+(N+2)时不发生进位的数的总个数。翻译过来就是个位数为0,1,2,非个位数可以为0,1,2,3且小于N的数有多少个。
这题开始时候感觉是数位dp,是有一种可取状态共两种状态的基础数位dp题,也搞了一发过了。
代码
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<ctype.h>
#include<algorithm>
#include<string>
#define PI acos(-1.0)
#define maxn 15
#define INF 1<<25
typedef long long ll;
using namespace std;
int len;
int nn[maxn];
int ans[maxn];
__int64 dfs(int pos,bool cmp)
{
if(pos==0)
return 1;
if(ans[pos]!=-1&&!cmp)
return ans[pos];
int news=(cmp==1?nn[pos]:9);
if(pos==1)
news=min(news,2);
else news=min(news,3);
__int64 aa=0;
for(int i=0;i<=news;i++)
{
bool c=(cmp&&(i==nn[pos]));
aa+=dfs(pos-1,c);
}
return cmp?aa:ans[pos]=aa;
}
int main()
{
__int64 num;
while(scanf("%I64d",&num)!=EOF)
{
num--;
memset(ans,-1,sizeof(ans));
int pp=0;
while(num)
{
nn[++pp]=num%10;
num/=10;
}
printf("%I64d\n",dfs(pp,1));
}
}
因为是数论专题里的题,所以还是要用组合数学的想法去思考,从给的数字的最高位往下搞,刚开始不是很理解"组合"是什么意思。
给出“组合数学”定义
狭义的组合数学主要研究满足一定条件的组态(也称组合模型)的存在、计数以及构造等方面的问题。 组合数学的主要内容有组合计数、组合设计、组合矩阵、组合优化(最佳组合)等。
这题的组合数学就是组合的问题。刚开始想要从高位向低位一位一位搞,每搞一位算出后面对应所有可取的数后向后跳一位,直到最后一位。
后来发现自己逗了,如果有某位数(最后一位除外)>=4时,即包括后面的所有情况,便需跳出循环,防止计算不应该计算的数。
代码
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<ctype.h>
#include<algorithm>
#include<string>
#define PI acos(-1.0)
#define maxn 15
#define INF 1<<25
typedef long long ll;
using namespace std;
int main()
{
__int64 tot;
char num[15];
while(scanf("%s",num)!=EOF)
{
__int64 ans=0;
int len=strlen(num);
for(int i=0; i<len; i++)
{
if((len-i-1))
{
if(num[i]>='4')
{
ans+=pow(4.0,len-1-i)*3;
break;
}
else ans+=pow(4.0,len-2-i)*3*(num[i]-'0');
}
else
{
if(num[i]>='3')
ans+=3;
else ans+=num[i]-'0';
}
}
printf("%I64d\n",ans); }
}

可能数据相对比较水的缘故,用两种算法提交都是15ms,不过用组合数学的方法明显时间复杂度比较低。

组合数学第一发 hdu 2451 Simple Addition Expression的更多相关文章

  1. HDU 2451 Simple Addition Expression(组合数学)

    主题链接:http://acm.hdu.edu.cn/showproblem.php?pid=2451 Problem Description A luxury yacht with 100 pass ...

  2. HDU 2451 Simple Addition Expression

    题目大意:有一个关于 简单加法表达式  的定义告诉你,就是  选一个数字i  如果 i+(i+1)+(i+2) 它的和,没有任何一位进位的话,那就是 一个i的简单加法表达式,求小于n的表达式数目. 题 ...

  3. hdu 2451 Simple Addition Expression(数位DP )成败在于细节

    亚洲区域赛的题,简单的数位DP题,注重细节. 任何细节都有可能导致wa,所以没有绝对的水题. 把握好细节,此题便A. #include<stdio.h> __int64 getans(__ ...

  4. HDU 2451 Simple Addition Expression(找规律,考验智商)

    题目 最近比赛的题目好多签到题都是找规律的考验智商的题目啊,,,我怎么越来越笨了,,,, 通过列举,可以发现规律: 从左往右按位扫这个数: 当数的长度大于1时: 当首位大于3时,答案就是4*4*4*… ...

  5. 【HDOJ】2451 Simple Addition Expression

    递推,但是要注意细节.题目的意思,就是求s(x) = i+(i+1)+(i+2),i<n.该表达中计算过程中CA恒为0(包括中间值)的情况.根据所求可推得.1-10: 31-100: 3*41- ...

  6. 【计数】Simple Addition Expression

    [来源] 2008年哈尔滨区域赛 [题目链接]: http://acm.hdu.edu.cn/showproblem.php?pid=2451 [参考博客]: HDU 2451 Simple Addi ...

  7. HDU2451:Simple Addition Expression

    Problem Description A luxury yacht with 100 passengers on board is sailing on the sea in the twiligh ...

  8. *HDU 2451 数学

    Simple Addition Expression Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Ja ...

  9. 10994 - Simple Addition(规律)

    Problem E Simple Addition Input: Standard Input Output: Standard Output Let’s define a simple recurs ...

随机推荐

  1. .Net程序员 Solr-5.3之旅 (一)Solr入门

    阅读目录 引言 Lunece是什么? Solr是什么 JAVA环境搭建 JAVA环境搭建之变量配置 Tomcat简单配置 结尾 引言 君子生非异也,善假于物也. Java和.Net哪个好,我们也不需要 ...

  2. IE浏览器中设为首页

    <a onclick="this.style.behavior='url(#default#homepage)';this.setHomePage('<%=Configurati ...

  3. .net判断用户使用的是移动设备还是PC

    using System.Text.RegularExpressions;//头部引入正则的命名空间 //为了加强准确性,防止支持wap的浏览器如opera,加入操作系统验证.openwave|后为p ...

  4. RAC检查各资源

  5. WPF 依赖属性与依赖对象

    在介绍依赖属性之前,我先介绍下属性的历史 属性的历史:      早期C++的类中,只有字段及方法,暴露数据靠的是方法, 但是字段直接暴露会不安全,所以才用方法来暴露,在设置的时候加些约束,在MFC中 ...

  6. Hibernate 多对多关联Demo

    以学生[Student ]与课程[Course ]之间的关系为例: //Course .java public class Course implements Serializable { priva ...

  7. SVN 不能提交, 看不到日志, 出现乱码. 解决方案.

    需要工具 sprite3: 点这里下载. 解决问题 如本文标题所写. 我遇到过几次一样的问题, 每次都很蛋疼的把目录重新检出, 浪费时间, 又伤了脾气. 下面是我在百度经验找到的一片帖子, 效果杠杠的 ...

  8. 【USACO 3.1.4】形成的区域

    [描述]         N个不同的颜色的不透明的长方形(1 <= N <= 1000)被放置在一张宽为A长为B的白纸上.这些长方形被放置时,保证了它们的边于白纸的边缘平行.所有的长方形都 ...

  9. js中callee与caller的区别

    callee是对象的一个属性,该属性是一个指针,指向参数arguments对象的函数首先我们来写个阶成函数:function chen(x){if (x<=1) {return 1;} else ...

  10. html5--canvas学习笔记

    1. 添加<canvas>元素 right: <canvas id="myCanvas" width="300" height="3 ...