805D - Minimum number of steps 思路:简单模拟,a每穿过后面一个b,b的个数+1,当这个a穿到最后,相当于把它后面的b的个数翻倍.每个a到达最后的步数相当于这个a与它后面已经到达最后的a之间的b的个数,只要从后面往前扫,记录b的个数,每遇到一个a,把b的个数加入答案,并且b的个数翻倍. 代码: #include<bits/stdc++.h> using namespace std; const int INF=0x3f3f3f3f; ; ; int main()…

http://codeforces.com/contest/805/problem/D [思路] 要使最后的字符串不出现ab字样,贪心的从后面开始更换ab为bba,并且字符串以"abbbb..."形式出现的话,那么需要替换的次数就是b的个数,并且b的个数会翻倍,因此遍历查找存在"ab”子串的位置,然后开始替换,并记录下每个位置开始及其后面b的个数,然后更新答案即可. [Accepted] #include <iostream> #include <stdio…

D. Minimum number of steps time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of s…

http://codeforces.com/contest/805/problem/D D. Minimum number of steps time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output We have a string of letters 'a' and 'b'. We want to perform some opera…

D. Minimum number of steps time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of s…

cf劲啊 原题: We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of substrings "ab" in the string and replace it with the string "bba". If we have no "ab" as a substring,…

D. Minimum number of steps time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of s…

[题目链接]:http://codeforces.com/contest/805/problem/D [题意] 给你一个字符串; 里面只包括a和b; 让你把里面的"ab"子串全都去掉; 方式是, 每次操作可以把"ab"替换成为"bba"; 直到整个字符串里面没有"ab"子串为止 [题解] 从ab开始 ab->bba 左边再加一个a的话 即aab 就相当于在bba左边加了一个a abba -> bbaba bbbba…

传送门:http://codeforces.com/contest/805/problem/D 对于一个由‘a’.‘b’组成的字符串,有如下操作:将字符串中的一个子串“ab”替换成“bba”.当字符串中不含有子串“ab”时,任务完成.求完成任务的最小操作次数(mod109+7). 最终,字符串的形式为: bbb...baaa...a 可以考虑寻找规律: a. i)“ab”→“bba”, ii)“aab”→“abba”→“bbaba”→“bbbbaa”, iii)“aaab”→“aabba”→“a…

题目链接:http://codeforces.com/contest/805/problem/D 题意:只有一个操作就是将ab变成bba直到不能变为止,问最少边几次. 题解:这题可以多列几组来找规律,事实上是可以递推的,递推可得到一个式子 a[n]=2*a[n-1]+1,然后化成通项公式. a[n]+1=2*(a[n-1]+1),a[n]+1=2^n,a[n]=2^n-1; 然后就可以解决b前面有几个a的问题了,ab=2^1-1,aab=2^2-1, #include <iostream> #…

最后肯定是bbbb...aaaa...这样. 你每进行一系列替换操作,相当于把一个a移动到右侧. 会增加一些b的数量……然后你统计一下就行.式子很简单. 喵喵喵,我分段统计的,用了等比数列……感觉智障.一个a一个a地统计答案即可. #include<cstdio> #include<cstring> #include<iostream> using namespace std; #define MOD 1000000007ll typedef long long ll;…

题意: 给定一串字符串,将所有“ab”的子串替换为“bba”,询问多少次操作后没有子串“ab”. 分析: 观察可得,将“ab”替换为“bba”有两种结果. ①a移到了b的后面 ②增加了一个b 而且最终的结果一定是前面全是b,后面全是a. 所以可以猜想从后往前数,设置一个B_cnt, 每当碰到一个b, 就b_cnt++, 碰到A, 就先加上一个b_cnt(因为每替换一次会将a移动后一格,所以要替换b_cnt次),再将b_cnt*2(多出b_cnt个b). #include <cstdio> #i…

思路: 找规律.参考了http://blog.csdn.net/harlow_cheng/article/details/71190999. 实现: #include <iostream> using namespace std; ; int main() { string s; cin >> s; int n = s.length(); ; ; ; i < n; i++) { % mod; ) % mod; } cout << sum << endl…

嗯... 题目链接:https://www.luogu.org/problemnew/show/CF804B 这道题没有什么技巧,只是一道找规律的题. 首先看到“ab”可以换成“bba”,所以首先要确定我们要逆序枚举(注意s从s_0开始),如果遇到a,则把ans += cntb,因为可以换的次数即为a后面b的个数,然后把cntb的个数乘二,因为一个ab可以换bba,也就一个b换两个b.如果遇到b,则直接cntb++即可,不需要别的操作. 注:mod可以随时模,不会影响最后结果. AC代码: #i…

题目如下: Given a m x n binary matrix mat. In one step, you can choose one cell and flip it and all the four neighbours of it if they exist (Flip is changing 1 to 0 and 0 to 1). A pair of cells are called neighboors if they share one edge. Return the min…

The Minimum Number of Variables 我们定义dp[ i ][ mask ]表示是否存在 处理完前 i 个a, b中存者 a存在的状态是mask 的情况. 然后用sosdp处理出,状态为state的a, 能组成的数字, 然后转移就好啦. #include<bits/stdc++.h> #define LL long long #define LD long double #define ull unsigned long long #define fi first #…

https://www.hackerrank.com/challenges/minimum-swaps-2/problem Minimum Swaps II You are given an unordered array consisting of consecutive integers  [1, 2, 3, ..., n] without any duplicates. You are allowed to swap any two elements. You need to find t…

There are a number of spherical balloons spread in two-dimensional space. For each balloon, provided input is the start and end coordinates of the horizontal diameter. Since it's horizontal, y-coordinates don't matter and hence the x-coordinates of s…

There are a number of spherical balloons spread in two-dimensional space. For each balloon, provided input is the start and end coordinates of the horizontal diameter. Since it's horizontal, y-coordinates don't matter and hence the x-coordinates of s…

Construct minimum number by reordering a given non-negative integer array. Arrange them such that they form the minimum number. Notice The result may be very large, so you need to return a string instead of an integer. Have you met this question in a…

There are a number of spherical balloons spread in two-dimensional space. For each balloon, provided input is the start and end coordinates of the horizontal diameter. Since it's horizontal, y-coordinates don't matter and hence the x-coordinates of s…

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2623 The number of steps Time Limit: 1000ms   Memory limit: 65536K  有疑问?点这里^_^ 题目描述 Mary stands in a strange maze, the maze looks like a triangle(the first layer have one room,the…

Given an integer array (index from 0 to n-1, where n is the size of this array), and an query list. Each query has two integers [start, end]. For each query, calculate the minimum number between index start and end in the given array, return the resu…

The number of steps Time Limit: 1000ms   Memory limit: 65536K  有疑问?点这里^_^ 题目描述 Mary stands in a strange maze, the maze looks like a triangle(the first layer have one room,the second layer have two rooms,the third layer have three rooms -). Now she st…

There are a number of spherical balloons spread in two-dimensional space. For each balloon, provided input is the start and end coordinates of the horizontal diameter. Since it's horizontal, y-coordinates don't matter and hence the x-coordinates of s…

Codeforces 55D Beautiful Number a positive integer number is beautiful if and only if it is divisible by each of its nonzero digits. Input The first line of the input contains the number of cases t (1 ≤ t ≤ 10). Each of the next t lines contains two…

The number of steps Time Limit: 1000ms   Memory limit: 65536K  有疑问?点这里^_^ 题目描写叙述 Mary stands in a strange maze, the maze looks like a triangle(the first layer have one room,the second layer have two rooms,the third layer have three rooms -). Now she…

转载请注明出处: http://www.cnblogs.com/fraud/          ——by fraud The number of steps Time Limit: 1 Sec  Memory Limit: 128 M Description Mary stands in a strange maze, the maze looks like a triangle(the first layer have one room,the second layer have two ro…

The number of steps nid=24#time" style="padding-bottom:0px; margin:0px; padding-left:0px; padding-right:0px; color:rgb(83,113,197); text-decoration:none; padding-top:0px"> Time Limit: 1000ms   Memory limit: 65536K  有疑问?点这里^_^ 题目描写叙述 Mary…

There are a number of spherical balloons spread in two-dimensional space. For each balloon, provided input is the start and end coordinates of the horizontal diameter. Since it's horizontal, y-coordinates don't matter and hence the x-coordinates of s…