UVA - 10706 Number Sequence

先找到是在哪个集合内,再找到是集合内的哪个元素,最后找到元素的第几位数

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
long long table[100010];
long long psum[100010];
int n=100000;
void maketable()
{
    int i,flag=1,x=0;
    for(i=1;i<=n;i++)
    {
        if(i%flag==0)
        {
            flag*=10;
            x++;
        }
        table[i]=table[i-1]+x;
        psum[i]+=table[i]+psum[i-1];
    }
}
int main()
{
    //freopen("in","r",stdin);
    //freopen("out","w",stdout);
    maketable();
    int T,i,t;
    long long x;
    char s[100];
    cin>>T;
    while(T--)
    {
        cin>>x;
        i=lower_bound(psum,psum+n,x)-psum;
        if(psum[i]==x)
            cout<<i%10<<endl;
        else
        {
            x-=psum[i-1];
            i=lower_bound(table,table+n,x)-table;
            if(table[i]==x)
                cout<<i%10<<endl;
            else
            {
                x-=table[i-1];
                sprintf(s,"%d",i);
                cout<<s[x-1]<<endl;
            }
        }
    }
    return 0;
}

UVA - 10706
Time Limit: 3000MS Memory Limit: Unknown 64bit IO Format: %lld & %llu

Submit Status

Description

Download as PDF

Problem B
Number Sequence
Input:
standard input
Output: standard output
Time Limit: 1 second

A single positive integer iis given. Write a program to find the digit located in the position iin the sequence of number groups S1S2…Sk. Each group Skconsists of a sequence of positive integer numbers ranging from 1 to k, written one after another. For example, the first 80 digits of the sequence are as follows:

11212312341234512345612345671234567812345678912345678910123456789101112345678910

Input

The first line of the input file contains a single integer t (1 <=t <=25), the number of test cases, followed by one line for each test case. The line for a test case contains the single integer i (1 <=i <=2147483647)

 

Output

There should be one output line per test case containing the digit located in the position i.

 

Sample Input                           Output for Sample Input

2

8

3

2

2


Problem source: Iranian Contest

Special Thanks: Shahriar Manzoor, EPS.

Source

Root :: AOAPC I: Beginning Algorithm Contests (Rujia Liu) :: Volume 4. Algorithm Design
Root :: Competitive Programming 2: This increases the lower bound of Programming Contests. Again (Steven & Felix Halim) :: Problem Solving Paradigms :: Divide and Conquer -Binary Search

Root :: Competitive Programming 3: The New Lower Bound of Programming Contests (Steven & Felix Halim) :: Problem Solving Paradigms :: Divide and Conquer :: Binary Search

内容概要:本文系统介绍了算术优化算法(AOA)的基本原理、核心思想及Python实现方法,并通过图像分割的实际案例展示了其应用价值。AOA是一种基于种群的元启发式算法,其核心思想来源于四则运算,利用乘除运算进行全局勘探,加减运算进行局部开发,通过数学优化器加速函数(MOA)和数学优化概率(MOP)动态控制搜索过程,在全局探索与局部开发之间实现平衡。文章详细解析了算法的初始化、勘探与开发阶段的更新策略,并提供了完整的Python代码实现,结合Rastrigin函数进行测试验证。进一步地,以Flask框架搭建前后端分离系统,将AOA应用于图像分割任务,展示了其在实际工程中的可行性与高效性。最后,通过收敛速度、寻优精度等指标评估算法性能,并提出自适应参数调整、模型优化和并行计算等改进策略。; 适合人群:具备一定Python编程基础和优化算法基础知识的高校学生、科研人员及工程技术人员,尤其适合从事人工智能、图像处理、智能优化等领域的从业者;; 使用场景及目标:①理解元启发式算法的设计思想与实现机制;②掌握AOA在函数优化、图像分割等实际问题中的建模与求解方法;③学习如何将优化算法集成到Web系统中实现工程化应用;④为算法性能评估与改进提供实践参考; 阅读建议:建议读者结合代码逐行调试,深入理解算法流程中MOA与MOP的作用机制,尝试在不同测试函数上运行算法以观察性能差异,并可进一步扩展图像分割模块,引入更复杂的预处理或后处理技术以提升分割效果。
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