Multiplication Puzzle Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7118   Accepted: 4385 Description The multiplication puzzle is played with a row of cards, each containing a single positive integer. During the move player takes one…

传送门:http://poj.org/problem?id=1651 Multiplication Puzzle Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13109   Accepted: 8034 Description The multiplication puzzle is played with a row of cards, each containing a single positive intege…

题目链接:http://poj.org/problem?id=1651 Description The multiplication puzzle is played with a row of cards, each containing a single positive integer. During the move player takes one card out of the row and scores the number of points equal to the prod…

Description The multiplication puzzle is played with a row of cards, each containing a single positive integer. During the move player takes one card out of the row and scores the number of points equal to the product of the number on the card taken…

Multiplication Puzzle Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 10010   Accepted: 6188 Description The multiplication puzzle is played with a row of cards, each containing a single positive integer. During the move player takes one…

题目链接:http://poj.org/problem? id=1651 题意:初使ans=0,每次消去一个值,位置在pos(pos!=1 && pos !=n) 同一时候ans+=a[pos-1]*a[pos]*a[pos+1].一直消元素直到最后剩余2个,求方案最小的ans是多少? 代码: #include <stdio.h> #include <ctime> #include <math.h> #include <limits.h> #…

题意: 给出一个序列,共n个正整数,要求将区间[2,n-1]全部删去,只剩下a[1]和a[n],也就是一共需要删除n-2个数字,但是每次只能删除一个数字,且会获得该数字与其旁边两个数字的积的分数,问最少可以获得多少分数? 思路: 类似于矩阵连乘的问题,用区间DP来做. 假设已知区间[i,k-1]和[k+1,j]各自完成删除所获得的最少分数,那么a[k]是区间a[i,j]内唯一剩下的一个数,那么删除该数字就会获得a[k]*a[i-1]*a[i+1]的分数了.在枚举k的时候要保证[i,j]的任一子区…

题目链接:id=1651">点击打开链 题意: 给定一个数组,每次能够选择内部的一个数 i 消除,获得的价值就是 a[i-1] * a[i] * a[i+1] 问最小价值 思路: dp[l,r] = min( dp[l, i] + dp[i, r] + a[l] * a[i] * a[r]); #include <cstdio> #include <iostream> #include <algorithm> #include <queue>…

题目链接:http://poj.org/problem?id=1651 思路:除了头尾两个数不能取之外,要求把所有的数取完,每取一个数都要花费这个数与相邻两个数乘积的代价,需要这个代价是最小的 用dp[i][j]表示区间[i,j]的最小代价,那么就有dp[i][j]=min(dp[i][k]+dp[k][j]+a[i]*a[k]*a[j]) i+1<=k<=j-1 #include<cstdio> #include<iostream> #include<algor…

The multiplication puzzle is played with a row of cards, each containing a single positive integer. During the move player takes one card out of the row and scores the number of points equal to the product of the number on the card taken and the numb…