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  • cf468A 24 Game

    A. 24 Game
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Little X used to play a card game called "24 Game", but recently he has found it too easy. So he invented a new game.

    Initially you have a sequence of n integers: 1, 2, ..., n. In a single step, you can pick two of them, let's denote them a and b, erase them from the sequence, and append to the sequence either a + b, or a - b, or a × b.

    After n - 1 steps there is only one number left. Can you make this number equal to 24?

    Input

    The first line contains a single integer n (1 ≤ n ≤ 105).

    Output

    If it's possible, print "YES" in the first line. Otherwise, print "NO" (without the quotes).

    If there is a way to obtain 24 as the result number, in the following n - 1 lines print the required operations an operation per line. Each operation should be in form: "a op b = c". Where a and b are the numbers you've picked at this operation; op is either "+", or "-", or "*";c is the result of corresponding operation. Note, that the absolute value of c mustn't be greater than 1018. The result of the last operation must be equal to 24. Separate operator sign and equality sign from numbers with spaces.

    If there are multiple valid answers, you may print any of them.

    Sample test(s)
    input
    1
    
    output
    NO
    
    input
    8
    
    output
    YES
    8 * 7 = 56
    6 * 5 = 30
    3 - 4 = -1
    1 - 2 = -1
    30 - -1 = 31
    56 - 31 = 25
    25 + -1 = 24

    其实就是sb题

    注意到每4个数可以消掉

    n<4直接打出NO

    n=4到7打表

    大于7的对4取mod之后各种乱搞

    #include<cstdio>
    #include<iostream>
    #include<cstring>
    #include<cstdlib>
    #include<cmath>
    #include<algorithm>
    #include<ctime>
    #include<set>
    #define LL long long 
    using namespace std;
    inline LL read()
    {
        LL x=0,f=1;char ch=getchar();
        while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
        while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
        return x*f;
    }
    int n;
    int main()
    {
    	scanf("%d",&n);
    	if (n<4){printf("NO
    ");return 0;}
    	int s;
    	if(n%4==0)printf("YES
    1 * 2 = 2
    2 * 3 = 6
    6 * 4 = 24
    "),s=n-4;
    	else if (n%4==1)printf("YES
    4 * 5 = 20
    20 + 2 = 22
    22 + 3 = 25
    25 - 1 = 24
    "),s=n-5;
    	else if (n%4==2)printf("YES
    5 * 6 = 30
    30 - 4 = 26
    26 - 3 = 23
    23 - 1 = 22
    22 + 2 = 24
    "),s=n-6;
    	else if (n%4==3)printf("YES
    7 + 6 = 13
    13 + 5 = 18
    18 + 4 = 22
    22 + 3 = 25
    25 - 2 = 23
    23 + 1 = 24
    "),s=n-7;
    	for(int j=n-s+1;j<=n;j+=2)
    	  printf("%d - %d = 1
    ",j+1,j);
    	for (int j=1;j<=s/4;j++)
    	  printf("1 - 1 = 0
    ");
    	for (int j=1;j<=s/4;j++)
    	  printf("24 + 0 = 24
    ");
    }
    

      

    ——by zhber,转载请注明来源
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  • 原文地址:https://www.cnblogs.com/zhber/p/4035911.html
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