Code Forces 469C 24 Game

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本文介绍了一种通过数学运算构造数字24的游戏算法。对于不同数量的初始整数，提供了不同的策略来达到目标数字24，包括直接构造和利用0进行构造的方法。

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A - A

Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 469C

Description

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 thema andb, erase them from the sequence, and append to the sequence eithera + b, ora - b, ora × b.

After n - 1 steps there is only one number left. Can you make this number equal to24?

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 followingn - 1 lines print the required operations an operation per line. Each operation should be in form: "aopb = c". Where a andb 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 ofc mustn't be greater than 10¹⁸. The result of the last operation must be equal to24. Separate operator sign and equality sign from numbers with spaces.

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

Sample Input

Input

Output

NO

Input

Output

YES
8 * 7 = 56
6 * 5 = 30
3 - 4 = -1
1 - 2 = -1
30 - -1 = 31
56 - 31 = 25
25 + -1 = 24

这题是个找规律的题，

当n<4的时候，一定是不可能构成24的,直接输出NO

n==4的时候，

 printf("1 * 2 = 2\n");
        printf("2 * 3 = 6\n");
        printf("6 * 4 = 24\n");

n>4的时候，构造n-(n-1)=1;1*24=24即可。

#include <stdio.h>
void solve(int n)
{
    if(n%2==0)
    {
        printf("1 * 2 = 2\n");
        printf("2 * 3 = 6\n");
        printf("6 * 4 = 24\n");
        while(n>5)
        {
            printf("%d - %d = 1\n",n,n-1);
            printf("1 * 24 = 24\n");
            n-=2;
        }
    }
    else
    {
        printf("5 - 1 = 4\n");
        printf("4 - 2 = 2\n");
        printf("2 * 3 = 6\n");
        printf("6 * 4 = 24\n");
        while(n>5)
        {
            printf("%d - %d = 1\n",n,n-1);
            printf("1 * 24 = 24\n");
            n-=2;
        }
    }
}
int main()
{
    int n;
    scanf("%d",&n);
    if(n<4)
        printf("NO\n");
    else
    {
        printf("YES\n");
        solve(n);
    }
    return 0;
}

出了构造n-(n-1)=1;

看了大牛的博客，还有一种方法是构造

  printf("0 * %d = 0\n", i);
        printf("0 + 24 = 24\n");

具体实现如下:

#include <stdio.h>
int main()
{
    int n;
    scanf("%d",&n);
    if(n<4)
        printf("NO\n");
    else if(n==4)
    {
        printf("YES\n");
        printf("1 * 2 = 2\n");
        printf("2 * 3 = 6\n");
        printf("6 * 4 = 24\n");
    }
    else if(n==5)
    {
        printf("YES\n");
        printf("5 - 2 = 3\n");
        printf("3 + 3 = 6\n");
        printf("6 * 1 = 6\n");
        printf("6 * 4 = 24\n");
    }
    else
    {
        printf("YES\n");
        printf("6 - 5 = 1\n");
        printf("1 - 1 = 0\n");
        printf("2 * 3 = 6\n");
        printf("6 * 4 = 24\n");
        for (int i = 7; i <= n; i++)
            printf("0 * %d = 0\n", i);
        printf("0 + 24 = 24\n");
    }
    return 0;
}

不过时间上，貌似构造n-(n-1)更快一点(31ms),0的构造是46ms.以此为鉴。