zoukankan      html  css  js  c++  java
  • Codeforces Round #347 (Div.2)_B. Rebus

    题目链接:http://codeforces.com/contest/664/problem/B

    B. Rebus
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    You are given a rebus of form ? + ? - ? + ? = n, consisting of only question marks, separated by arithmetic operation '+' and '-', equality and positive integer n. The goal is to replace each question mark with some positive integer from 1 to n, such that equality holds.

    Input

    The only line of the input contains a rebus. It's guaranteed that it contains no more than 100 question marks, integer n is positive and doesn't exceed 1 000 000, all letters and integers are separated by spaces, arithmetic operations are located only between question marks.

    Output

    The first line of the output should contain "Possible" (without quotes) if rebus has a solution and "Impossible" (without quotes) otherwise.

    If the answer exists, the second line should contain any valid rebus with question marks replaced by integers from 1 to n. Follow the format given in the samples.

    Examples
    input
    ? + ? - ? + ? + ? = 42
    output
    Possible
    9 + 13 - 39 + 28 + 31 = 42
    input
    ? - ? = 1
    output
    Impossible
    input
    ? = 1000000
    output
    Possible
    1000000 = 1000000

    解题报告:

    看到大神们的代码,简直碉堡了。

    1、这里的输入要掌控好。

    2、思路:先置每个位置为1,记录下加上的数的个数sa,减去的个数sb,这样,p(离目标结果)只差p-n了。

    3、根据每个问号前面的符号,自增,达到结果。

    #include <cstdio>
    
    using namespace std;
    
    int a[1005];    ///结果
    int v[1005];    ///记录每个符号
    
    char c[2];  ///符号命令
    
    int main()
    {
        scanf("%s",c);  ///输入问号
        scanf("%s",c);  ///输入符号
        int m=1;    ///?的个数,即要填的数字的个数
        int sa=1;   ///加数的个数
        int sb=0;   ///减数的个数
        v[1]=1;
    
        while(c[0]!='=')
        {
            if(c[0]=='+')
            {
                v[++m]=1;
                sa++;
            }
            else {
                v[++m]=-1;
                sb++;
            }
            scanf("%s",c);///输入问号
            scanf("%s",c);///输入符号
        }
    
        int n;
        scanf("%d",&n); ///目标结果
    
        for(int i=1;i<=m;i++)   ///全部填上1
            a[i]=1;
    
        int p=sa-sb;
    
        for(int i=1;i<=m;i++)   ///一个一个处理各个数
        {
            while((p<n)&&(v[i]==1)&&(a[i]<n))
                a[i]++,p++;
            while((p>n)&&(v[i]==-1)&&(a[i]<n))
                a[i]++,p--;
        }
    
        if(p!=n)
        {
            printf("Impossible
    ");
            return 0;
        }
        printf("Possible
    ");
        printf("%d ",a[1]);
    
        for(int i=2;i<=m;i++)
        {
            if(v[i]==1) printf("+ ");
            else printf("- ");
            printf("%d ",a[i]);
        }
        printf("= %d
    ",n);
    }
  • 相关阅读:
    SpringBoot整合ActiveMQ同时支持P2P和发布订阅模式(三)
    SpringBoot整合ActiveMQ的publish/subscribe发布订阅模式(二)
    Windows启动ActiveMQ报Wrapper Stopped错误
    IDEA从远程仓库克隆项目
    Git的安装
    IDEA上传项目到使用github上
    Mybaits的逆向工程
    posman测试接口需要登录验证的使用
    SSM整合SpringSecurity
    SpringBoot整合MongoDB的连接用户名和密码问题
  • 原文地址:https://www.cnblogs.com/TreeDream/p/5406024.html
Copyright © 2011-2022 走看看