Magic Trick（水题）

最新推荐文章于 2023-04-10 19:01:00 发布

Baiyi_destroyer

最新推荐文章于 2023-04-10 19:01:00 发布

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分类专栏：水题集

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本文介绍了一个基于数学魔术的编程挑战，任务是通过一系列算术操作判断哪些初始整数会导致过程出现分数或负数，最终统计数量。代码使用C++实现，逐个检查1到100之间的整数。

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Description

Your friend has come up with a math trick that supposedly will blow your mind. Intrigued, youask your friend to explain the trick.

First, you generate a random positive integer k between 1 and 100. Then, your friend will giveyou n operations to execute. An operation consists of one of the four arithmetic operations ADD,SUBTRACT, MULTIPLY, or DIVIDE, along with an integer-valued operand x. You are supposed toperform the requested operations in order.

You don’t like dealing with fractions or negative numbers though, so if during the process, theoperations generate a fraction or a negative number, you will tell your friend that he messed up.

Now, you know the n operations your friend will give. How many of the first 100 positiveintegers will cause your friend to mess up?

Input

The first line of input contains a single positive integer n (1 ≤ n ≤ 10). Each of the next n linesconsists of an operation, followed by an operand. The operation is one of the strings ADD, SUBTRACT,MULTIPLY, or DIVIDE. Operands are positive integes not exceeding 5.

Output

Print, on a single line, a single integer indicating how many of the first 100 positive integers willresult in you telling your friend that he messed up.

Sample Input

SUBTRACT 5

Sample Output

Sample Input

DIVIDE 2

Sample Output

Sample Input

ADD 5

DIVIDE 5

Sample Output

题解：当时做的时候没理解题意，一直w，后来一看，真简单......，就是每一步执行都把小数和负数排除，最后求和。

代码如下：

#include <iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<stack>
#include<queue>
#include<deque>
using namespace std;
int main()
{
    int t,b,count=0,c[200];
    char a[20];
    cin>>t;
    for(int i=1; i<=100; i++)
        c[i]=i;
    while(t--)
    {
        cin>>a>>b;
        if(a[0]=='A')
            for(int i=1; i<=100; i++)
                if(c[i]!=-0x3f3f3f3f)
                    c[i]+=b;
        if(a[0]=='S')
            for(int i=1; i<=100; i++)
                if(c[i]!=-0x3f3f3f3f)
                    c[i]-=b;
        if(a[0]=='M')
            for(int i=1; i<=100; i++)
                if(c[i]!=-0x3f3f3f3f)
                    c[i]*=b;
        if(a[0]=='D')
            for(int i=1; i<=100; i++)
                if(c[i]!=-0x3f3f3f3f)
                {
                    if(c[i]%b==0)
                        c[i]/=b;
                    else
                        c[i]=-99999;
                }
        for(int i=1; i<=100; i++)
            if(c[i]<0&&c[i]!=-0x3f3f3f3f)
            {
                c[i]=-0x3f3f3f3f;
                count++;
            }
    }
    printf("%d\n",count);
    return 0;
}