http://codeforces.com/problemset/problem/204/A 题意:给定一个[L,R]区间,求这个区间里面首位和末尾相同的数字有多少个 思路:考虑这个问题满足区间加减,我们只考虑[1,n],考虑位数小于n的位数的时候,我们枚举头尾的数是多少,然后乘上10的某幂次,再考虑位数相等时,从高位往低位走,先考虑头尾数字小于最高位的情况,也像刚才那个随便取,当头尾数字等于最高位时,从高往低走,先算不与这位相等的,走下一步就代表与这位相等.最后要记得判断一下如果原数字头等于尾…

题目描述: Little Elephant and Interval time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output The Little Elephant very much loves sums on intervals. This time he has a pair of integers l and r (l ≤ r…

CodeForces - 204C Little Elephant and Furik and Rubik 个人感觉是很好的一道题 这道题乍一看我们无从下手,那我们就先想想怎么打暴力 暴力还不简单?枚举所有字串,再枚举所有位置,算出所有答案不就行了 我们自然不能无脑暴力,但是暴力可以给我们启发 我们知道所有对答案做出贡献的字符一定是相同的(废话) 所以我们可以O(n^2)首先枚举两个字符串中相同的字符然后再考虑如何贡献 然后计算出所有的方案下的值,再除以n*(n+1)*(2*n+1)/6 [不知…

题意: 有一种个位数与最高位数字相等的数字,求在l,r的范围内,这样的数字的个数. 思路: 找下规律就知道当当n>10的时候除去个位以后的答案等于n/10,然后考虑第一个数字是否小于最后一个.小于减一,还要加上个位一定存在的9位数 import java.util.Scanner; public class xxz { public static void main(String[] args) { Scanner sc = new Scanner(System.in); long l = sc…

 Little Elephant and Chess Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Submit Status Practice CodeForces 259A Description The Little Elephant loves chess very much. One day the Little Elephant and his friend decided…

Little Elephant and Array Time Limit: 4000ms Memory Limit: 262144KB This problem will be judged on CodeForces. Original ID: 221D64-bit integer IO format: %I64d      Java class name: (Any) The Little Elephant loves playing with arrays. He has array a,…

The Little Elephant very much loves sums on intervals. This time he has a pair of integers l and r (l ≤ r). The Little Elephant has to find the number of such integers x (l ≤ x ≤ r), that the first digit of integer x equals the last one (in decimal n…

This problem can be solve in simpler O(NsqrtN) solution, but I will describe O(NlogN) one. We will solve this problem in offline. For each x (0 ≤ x < n) we should keep all the queries that end in x. Iterate that x from 0 to n - 1. Also we need to kee…

Little Elephant and Broken Sorting 怎么感觉这个状态好难想到啊.. dp[ i ][ j ]表示第 i 个数字比第 j 个数字大的概率.转移好像比较显然. #include<bits/stdc++.h> #define LL long long #define LD long double #define ull unsigned long long #define fi first #define se second #define mk make_pair…

Little Elephant and LCM #include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define PLL pair<LL, LL> #define PLI pair<LL, int> #define PII pair<int, int> #define SZ(x) ((int)x.size()…