一.题目大意 有这样一个序列包含S1,S2,S3...SK,每一个Si包括整数1到 i.求在这个序列中给定的整数n为下标的数. 例如,前80位为11212312341234512345612345671234567812345678912345678910123456789101112345678910,第8位为2. 二.题解 动手做这道题之前要了解所求的是第n位而不是某一个Si中的第几个数,一位数占一位,两位数占两位,三位占三位...由于每一组都比前一组多一个数,如果这个数是一位数,则该组数的…

题目地址:http://poj.org/problem?id=1019 Number Sequence Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 35680   Accepted: 10287 Description A single positive integer i is given. Write a program to find the digit located in the position i in…

Number Sequence Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 36013   Accepted: 10409 Description A single positive integer i is given. Write a program to find the digit located in the position i in the sequence of number groups S1S2..…

找规律,先找属于第几个循环,再找属于第几个数的第几位...... Number Sequence Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 31552 Accepted: 8963 Description A single positive integer i is given. Write a program to find the digit located in the position i in the sequ…

这是一个看似简单,其实很难受. 本来我想发挥它的标题轨道基础.没想到反被消遣-_-|||. 看它在个人基础上,良好的数学就干脆点,但由于过于频繁,需求将被纳入全,因此,应该难度4星以上. 方法就是直接打表.然后直接模拟.利用打表去掉一大段数据,剩下数据量十分小了.故此能够直接模拟. 打表是为了计算前面的周期数,把周期数直接去掉. 主要难点是后面10位数以上的数有2位, 3位,4位等情况要考虑.- 以下使用getNewNums一个函数攻克了,想通了,就几行代码,还不用难理解的数学计算,呵呵. 然后…

https://vjudge.net/problem/POJ-1019 题意 给一串1 12 123 1234 12345 123456 1234567 12345678 123456789 12345678910 1234567891011这种形式的串,问这个串的第i个位置的数字是什么. 分析 这道题的重点在于到第k组时应该为几位数,即对于某个数x,它应该为几位数.答案是log10(x)+1.这样剩下的便是打表预处理了,找出每组的起始位置,再处理出一个最长的组. #include<iostre…

POJ 3273 Monthly Expense二分查找(最大值最小化问题) 题目:Monthly Expense Description Farmer John is an astounding accounting wizard and has realized he might run out of money to run the farm. He has already calculated and recorded the exact amount of money (1 ≤ mon…

POJ 3273 Monthly Expense 此题与POJ3258有点类似,一开始把判断条件写错了,wa了两次,二分查找可以有以下两种: ){ mid=(lb+ub)/; if(C(mid)<=m) ub=mid; ; //此时下限过小 } out(ub);//out(lb) 我一开始是写的下面这种,下面这种要单独判断lb和ub的值,因为用下面这种判断lb,ub都可能成立 ){ mid=(lb+ub)/; if(C(mid)<=m) ub=mid; else lb=mid; } if(C(…

题目描述 Given a list of N integers with absolute values no larger than 10 15, find a non empty subset of these numbers which minimizes the absolute value of the sum of its elements. In case there are multiple subsets, choose the one with fewer elements.…

#include<cstdio> #include<iostream> #include<cstring> #include<algorithm> #define N 2000 #define M 1000010 #define inf 1<<30 using namespace std; struct Edge{ int to,val,next; }edge[M]; ]; void addedge(int from,int to,int val…