zoukankan      html  css  js  c++  java
  • Quicksum-S.B.S.

    quicksum

    Queation:

    Given a string of digits, find the minimum number of additions required for the string to equal some target number. Each addition is the equivalent of inserting a plus sign somewhere into the string of digits. After all plus signs are inserted, evaluate the sum as usual. For example, consider the string "12" (quotes for clarity). With zero additions, we can achieve the number 12. If we insert one plus sign into the string, we get "1+2", which evaluates to 3. So, in that case, given "12", a minimum of 1 addition is required to get the number 3. As another example, consider "303" and a target sum of 6. The best strategy is not "3+0+3", but "3+03". You can do this because leading zeros do not change the result.

    Write a class QuickSums that contains the method minSums, which takes a String numbers and an int sum. The method should calculate and return the minimum number of additions required to create an expression from numbers that evaluates to sum. If this is impossible, return -1.

    example:

    "382834"

    100

    Returns: 2

    There are 3 ways to get 100. They are 38+28+34, 3+8+2+83+4 and 3+82+8+3+4. The minimum required is 2.

    Constraints

    -      numbers will contain between 1 and 10 characters, inclusive.

    -      Each character in numbers will be a digit.

    -      sum will be between 0 and 100, inclusive.

    -   the string will be shorter than 100 bit.

    ---------------------------------------------我是分割线--------------------------------------------------------------

    本题有多种方法,例如“记忆化搜索+剪枝”,但我用的是DP。

    由于数据太弱(加号数小于10,和不大于100……)所以开个三维数组dp[i][j][k]。

    i、j表示字符串从i开始到j表示的数;

    k表示此时和为k;

    数组内存放所需加号数。

    不多说,上代码:

     1 #include<iostream>
     2 #include<cstdio>
     3 #include<cstring>
     4 #include<queue>
     5 #include<cmath>
     6 #include<algorithm>
     7 #include<cstdlib>
     8 using namespace std;
     9 int cut(string,int,int);
    10 int read(){
    11     int x=0,f=1;char ch=getchar();
    12     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    13     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    14     return x*f;
    15 }
    16 long long chang=0;
    17 int dp[101][101][101];
    18 int main()
    19 {
    20     string s;long long sum;long long ans=0;
    21     cin>>s;
    22     chang=s.length();
    23     cin>>sum;
    24     for(int i=0;i<=100;i++)
    25      for(int j=0;j<=100;j++)
    26       for(int k=0;k<=100;k++)
    27           dp[i][j][k]=11;
    28 //    cout<<dp[0][chang-1][sum]<<endl;
    29     for(int i=0;i<chang;i++)
    30      for(int j=0;i+j<chang;j++)
    31      {
    32         long long num=cut(s,i,i+j);
    33         if(num<=sum) dp[i][i+j][num]=0;
    34      } 
    35 //    cout<<dp[0][chang-1][sum]<<endl; 
    36     for(int i=1;i<chang;i++)                 //数长 
    37      for(int head=0;head+i<chang;head++)     //始位 
    38       for(int j=0;j<=sum;j++)                //
    39        for(int k=head;k<head+i;k++)          //加号位 
    40         for(int ss=0;j-ss>0;ss++)            //中间和 
    41         dp[head][head+i][j]=min((dp[head][k][j-ss]+dp[k+1][head+i][ss])+1,dp[head][head+i][j]);
    42     ans=dp[0][chang-1][sum];
    43     if(ans==11) ans=-1;
    44     cout<<ans;
    45     return 0;    
    46 }
    47 int cut(string s,int a,int b)
    48 {
    49     long long n=0;
    50     for(int i=a;i<=b;i++)
    51        n=n*10+(s[i]-'0');
    52     return n;   
    53 }
  • 相关阅读:
    久违的问候-----eclipse中搭建maven项目2016年
    jdbc在mysql下一次执行多条sql脚本
    Oracle客户端连接远程Oracle服务中文乱码问题
    Hibernate 3.3.2 文档翻译 Day01
    Linux学习之Exam系统发布
    js封装用户选项传递给Servlet之考试系统二
    MySQL中的全文索引
    60分钟Python快速学习(给发哥一个交代)
    优化MySchool数据库设计之【巅峰对决】
    微冷的雨Java基础学习手记(一)
  • 原文地址:https://www.cnblogs.com/AwesomeOrion/p/5356273.html
Copyright © 2011-2022 走看看