1 second
256 megabytes
standard input
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?
The first line contains a single integer n (1 ≤ n ≤ 105).
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.
1
NO
8
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\n");return 0;}
int s;
if(n%4==0)printf("YES\n1 * 2 = 2\n2 * 3 = 6\n6 * 4 = 24\n"),s=n-4;
else if (n%4==1)printf("YES\n4 * 5 = 20\n20 + 2 = 22\n22 + 3 = 25\n25 - 1 = 24\n"),s=n-5;
else if (n%4==2)printf("YES\n5 * 6 = 30\n30 - 4 = 26\n26 - 3 = 23\n23 - 1 = 22\n22 + 2 = 24\n"),s=n-6;
else if (n%4==3)printf("YES\n7 + 6 = 13\n13 + 5 = 18\n18 + 4 = 22\n22 + 3 = 25\n25 - 2 = 23\n23 + 1 = 24\n"),s=n-7;
for(int j=n-s+1;j<=n;j+=2)
printf("%d - %d = 1\n",j+1,j);
for (int j=1;j<=s/4;j++)
printf("1 - 1 = 0\n");
for (int j=1;j<=s/4;j++)
printf("24 + 0 = 24\n");
}