zoukankan      html  css  js  c++  java
  • POJ 1651 Multiplication Puzzle

    传送门@百度

    Multiplication Puzzle
    Time Limit: 1000MS   Memory Limit: 65536K
         

    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 and the numbers on the cards on the left and on the right of it. It is not allowed to take out the first and the last card in the row. After the final move, only two cards are left in the row.

    The goal is to take cards in such order as to minimize the total number of scored points.

    For example, if cards in the row contain numbers 10 1 50 20 5, player might take a card with 1, then 20 and 50, scoring
    10*1*50 + 50*20*5 + 10*50*5 = 500+5000+2500 = 8000

    If he would take the cards in the opposite order, i.e. 50, then 20, then 1, the score would be
    1*50*20 + 1*20*5 + 10*1*5 = 1000+100+50 = 1150.

    Input

    The first line of the input contains the number of cards N (3 <= N <= 100). The second line contains N integers in the range from 1 to 100, separated by spaces.

    Output

    Output must contain a single integer - the minimal score.

    Sample Input

    6
    10 1 50 50 20 5
    

    Sample Output

    3650

    Source

    Northeastern Europe 2001, Far-Eastern Subregion
     
    水dp
     
     1 #include<set>
     2 #include<cstdio>
     3 #include<cstdlib>
     4 #include<cstring>
     5 #include<iostream>
     6 #include<algorithm>
     7 using namespace std;
     8 const int N = 110;
     9 #define For(i,n) for(int i=1;i<=n;i++)
    10 #define Rep(i,l,r) for(int i=l;i<=r;i++)
    11 #define Down(i,r,l) for(int i=r;i>=l;i--)
    12 int n,A[N];
    13 long long dp[N][N];
    14 
    15 int main(){
    16     scanf("%d",&n);
    17     For(i,n) scanf("%d",&A[i]);
    18     For(i,n-2) dp[i][i+2]=A[i]*A[i+2]*A[i+1];
    19     Down(i,n-2,1)
    20       Rep(j,i+2,n)
    21         Rep(k,i+1,j-1)
    22           if(dp[i][j])
    23                 dp[i][j]=min(dp[i][j],dp[i][k]+dp[k][j]+A[i]*A[k]*A[j]);
    24           else  dp[i][j]=dp[i][k]+dp[k][j]+A[i]*A[k]*A[j];
    25     cout<<dp[1][n]<<endl;   
    26     return 0;
    27 }
  • 相关阅读:
    曲面的外在几何(一)---曲面的基本理论
    几个积性函数的均值
    多重小数部分和的渐近式与小数部分积分(Ⅱ)
    二重小数部分和的渐近式
    一个极限问题
    正整数互素的概率问题
    多重小数部分和的渐近式与Ovidiu Furdui积分问题
    $prodlimits_{substack{(k,n)=1 \ 1leqslant k leqslant n}} k$ 的阶
    2016 年中国科学院大学数学分析考研试题
    无平方因子数的分布 (Ⅰ)
  • 原文地址:https://www.cnblogs.com/zjdx1998/p/4053785.html
Copyright © 2011-2022 走看看