【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...
【Number OfWA】 0 【完整代码】 #include<bits/stdc++.h>usingnamespacestd;#definelsonl,m,rt<<1#definersonm+1,r,rt<<1|1#defineLLlonglong#definerep1(i,a,b)for(inti=a;i<=b;i++)#definerep2(i,a,b)for(inti=a;i>=b;i--)#definempmake_pair#definepbpush_back#definefifirst#define...
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...
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, our job is done. Print the minimum number of steps we ...
Print the minimum number of steps modulo109 + 7. Samples 输入数据 1 ab Copy 输出数据 1 1 Copy 输入数据 2 aab Copy 输出数据 2 3 Copy Note The first example: "ab" → "bba". The second example: "aab" → "abba" → "bbaba" → "bbbbaa"....
3) divide A by one of it's prime factor. Find minimum number of steps to reduce A to 1. 1) it is best to divide by max prime factor or finding nearest prime number whichever is minimum. This is codeforces problem but I can't find it...
Given two strings s1 and s2, the task is to find the minimum number of steps required to convert s1 into s2. The only operation allowed is to swap adjacent elements in the first string. Every swap is counted as a single step. Examples: Input: s1 = “abcd”, s2 = “cdab” Output:...
codeforces 805 D. Minimum number of steps(数学) 2017-05-05 08:33 −... Gealo 0 265 LeetCode 452. Minimum Number of Arrows to Burst Balloons 2019-12-19 11:56 −原题链接在这里:https://leetcode.com/problems/minimum-number-of-arrows-to-burst-balloons/ 题目: There are a number of...
Codeforces805D. Minimum number of steps 题目意思就是给你一个字符串然后每个ab要变bba,然后问你最少变化多少次。 不管怎么变最后的字符串都是变成了bbbbbb...aaaaaa这种形式,a移到了最右,b移到了最左,可以知道不管怎么变,变化次数都是一样的,因为每个a后面直接相连的b的个数就是其对应的变化次数。所以问题...
Codeforces 805D - Minimum number of steps 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...