1443:[例题4]Addition Chains 题解 注释在代码里 注意优化搜索顺序以及最优化剪枝 代码 #include<iostream> #include<cstdio> #include<algorithm> #include<cmath> #include<cstring> #include<string> #include<queue> #include<functional> using na…

  Addition Chains  An addition chain for n is an integer sequence  with the following four properties: a0 = 1 am = n a0<a1<a2<...<am-1<am For each k ( ) there exist two (not neccessarily different) integers i and j ( ) with ak =ai +aj You a…

Addition Chains Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5454   Accepted: 2923   Special Judge Description An addition chain for n is an integer sequence <a0, a1,a2,...,am="">with the following four properties: a0 = 1 a…

[题目描述] An addition chain for n is an integer sequence with the following four properties: a0 = 1 am = n a0 < a1 < a2 < ... < am-1 < am For each k (1<=k<=m) there exist two (not necessarily different) integers i and j (0<=i, j<=k…

Addition Chains 题面 对于一个数列 \(a_1,a_2 \dots a_{m-1},a_m\) 且 \(a_1<a_2 \dots a_{m-1}<a_m\). 数列中的一个数 \(a_k(2<k<=m)\) ,都有两个数 \(a_i,a_j(1<=i,j<k)\) 满足 \(a_i+a_j=a_k\)( \(i\) 可以等于\(j\) ). 换句话说就是 \(a_k\) 前面有两个数可以加起来等于 \(a_k\)​ .这种数列就是加法链. 题目输入一个…

又是一道剪枝剪了半天的搜索题...题目传送 要充分利用题目中的约束条件:1.:2.对于每个k(1≤k≤m)k(1≤k≤m)满足ak=ai+aj(0≤i,j≤k−1)ak=ai+aj(0≤i,j≤k−1),这里i与j可以相等.由此可以推出a1一定=2(也能减少很多操作次数了吧) 还是先找找搜索过程要面临的状态和有关维度:目标数n,答案序列的长度limit,序列中的每个数ai,当前序列中的最后一个数last. 由于如果不对序列长度加以限制,普通的深搜便会“一发不可收拾”,所以这里用迭代加深搜索. 可…

Description An addition chain for n is an integer sequence  with the following four properties: a0 = 1 am = n a0<a1<a2<...<am-1<am For each k ( ) there exist two (not neccessarily different) integers i and j ( ) with ak =ai +aj You are give…

Description An addition chain for n is an integer sequence with the following four properties: a0 = 1 am = n a0 < a1 < a2 < ... < am-1 < am For each k (1<=k<=m) there exist two (not necessarily different) integers i and j (0<=i, j&…

An addition chain for n is an integer sequence <a0, a1,a2,...,am=""> with the following four properties: a0 = 1 am = n a0 < a1 < a2 < ... < am-1 < am For each k (1<=k<=m) there exist two (not necessarily different) int…

此题不难,主要思路便是IDDFS(迭代加深搜索),关键在于优化. 一个IDDFS的简单介绍,没有了解的同学可以看看: https://www.cnblogs.com/MisakaMKT/articles/10767945.html 我们可以这么想,设当前规定长度为M,题目要求得出的数为N. 在搜索中,当前的步数为step,当前的数列为 数组a. 首先来确定思路,便是在以得出的数列a中枚举每两个数相加得出sum,然后继续搜索下一步. 初步的代码便是: void iddfs(int step) {…