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最新推荐文章于 2025-09-11 23:50:21 发布

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最新推荐文章于 2025-09-11 23:50:21 发布

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文章标签： java

原文链接：http://www.cnblogs.com/WQHui/p/7575166.html

本文介绍了一个关于古董打印机成本优化的问题，主人公Zero拥有一台老式打印机，该打印机无法长时间工作且会磨损。为了完美地打印一篇包含N个单词的文章，每个单词都有一个打印成本Ci，每行打印k个单词会有固定成本Mi。文章探讨了如何计算最小化总打印成本的方法。

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Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)

Problem Description

Zero has an old printer that doesn't work well sometimes. As it is antique, he still like to use it to print articles. But it is too old to work for a long time and it will certainly wear and tear, so Zero use a cost to evaluate this degree.
One day Zero want to print an article which has N words, and each word i has a cost Ci to be printed. Also, Zero know that print k words in one line will cost

M is a const number.
Now Zero want to know the minimum cost in order to arrange the article perfectly.

Input

There are many test cases. For each test case, There are two numbers N and M in the first line (0 ≤ n ≤ 500000, 0 ≤ M ≤ 1000). Then, there are N numbers in the next 2 to N + 1 lines. Input are terminated by EOF.

Output

A single number, meaning the mininum cost to print the article.

Sample Input

5 5 5 9 5 7 5

Sample Output

230

Author

Xnozero

Source

2010 ACM-ICPC Multi-University Training Contest（7）——Host by HIT

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Code：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#define INF 100000000000000008
using namespace std;
long long f[500008],sum[500008];
int n,m;

int main(){
    while(scanf("%d%d",&n,&m)==2){
    for(int i=1;i<=n;i++)
        f[i]=INF;
    for(int i=1;i<=n;i++){
        int x;
        scanf("%d",&x);
        sum[i]=sum[i-1]+x;
    }
    int flag=0,k=0;
    for(int i=1;i<=n;i++){
        for(int j=flag;j<=i;j++)
            if(f[i]>f[j]+(sum[i]-sum[j])*(sum[i]-sum[j])+m){
                k=j;
                f[i]=f[j]+(sum[i]-sum[j])*(sum[i]-sum[j])+m;
            }
        flag=k;
    }
    printf("%lld\n",f[n]);
}
}