zoukankan      html  css  js  c++  java
  • uav 11258 String Partition (DP)

    Problem F - String Partition
                                                                                                                        Time limit: 3.000 seconds

    John was absurdly busy for preparing a programming contest recently. He wanted to create a ridiculously easy problem for the contest. His problem was not only easy, but also boring: Given a list of non-negative integers, what is the sum of them?



    However, he made a very typical mistake when he wrote a program to generate the input data for his problem. He forgot to print out spaces to separate the list of integers. John quickly realized his mistake after looking at the generated input file because each line is simply a string of digits instead of a list of integers.

    He then got a better idea to make his problem a little more interesting: There are many ways to split a string of digits into a list of non-zero-leading (0 itself is allowed) 32-bit signed integers. What is the maximum sum of the resultant integers if the string is split appropriately?

    Input
    The input begins with an integer N ( 500) which indicates the number of test cases followed. Each of the following test cases consists of a string of at most 200 digits.

    Output
    For each input, print out required answer in a single line.

    Sample input

    6
    1234554321
    5432112345
    000
    121212121212
    2147483648
    11111111111111111111111111111111111111111111111111111

    Sample output
    1234554321
    543211239
    0
    2121212124
    214748372
    5555555666
     
     
    题意:给定一个字符串序列。让你把它划分成几个int型的数字,而且使这些数字之和最大。

     
    思路:dp[i]表示前i个字符划分时的和的最大值。
    dp[i] = max(dp[i], dp[i-j]+num),num为从后往前划分为j长度的数字。

     
     
     
    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <cmath>
    #include <string>
    #include <climits>
    #define LL long long
    using namespace std;
    const LL inf=INT_MAX;
    const int maxn=210;
    
    LL dp[maxn];
    string str;
    
    void initial()
    {
        memset(dp,0,sizeof(dp));
    }
    
    void input()
    {
        cin>>str;
    }
    
    int get_num(char ch)
    {
        return ch-'0';
    }
    
    void solve()
    {
        int len=str.size();
        dp[1]=get_num(str[0]);
    
        for(int i=2;i<=len;i++)
        {
             LL tmp=1,now=0;
             for(int k=i-1,j=1;k>=0;k--,j++)
             {
                  now=get_num(str[k])*tmp+now;
                  if(now>inf)  break;
                  dp[i]=max(dp[i],dp[i-j]+now);
                  tmp*=10;
             }
        }
        printf("%lld
    ",dp[len]);
    }
    
    int main()
    {
        int T;
        scanf("%d",&T);
        while(T--)
        {
             initial();
             input();
             solve();
        }
        return 0;
    }
    

  • 相关阅读:
    表单验证总结
    <wp8>_______环境搭建
    <二维码>———二维码生成器之绘制二维码
    <图片>———屏幕截图、图片保存至图片库
    《ListBox》———实现分页追加效果
    <wp7>———Zip解压缩
    <Toolkit>———LockablePivot
    <div>设置宽度,汉字正常换行,输入字母/数字不换行的解决方案分析
    <wp7查看独立存储工具>———2012年11月后仍可以工具
    wp7丿____在 Windows Phone 中如何测试与照片选择器或相机启动器交互的应用
  • 原文地址:https://www.cnblogs.com/mengfanrong/p/5097911.html
Copyright © 2011-2022 走看看