CodeForces 468A A. 24 Game

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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 "*"; cis 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.

Examples
input
1
output
NO
input
8
output
YES8 * 7 = 566 * 5 = 303 - 4 = -11 - 2 = -130 - -1 = 3156 - 31 = 2525 + -1 = 24

题目大意：给你一个n，让你用一个1到n的每一个数字在n-1步内利用加减乘运算得到24，每一个式子得出来的答案可以用，式子左边的数字用完了就没了，如果可以，输出YES和每一步，否则输出NO。

乍一看这道题感觉很难，其实很水，在4之前都是得不出答案的，大于等于4之后就可以了，那么怎样写出这个式子呢，其实也不难，固定几个式子可以得到24的，然后将多出来的数字相减得1，再循环24*1即可

#include<iostream>#include<cstdio>using namespace std;int main(){    int n;    while(cin>>n){        if(n<4) cout<<"NO"<<endl;        else{        cout<<"YES"<<endl;        int i=n;            if(i%2==0){                while(i>4){                    printf("%d - %d = %d\n",i,i-1,1);                    i-=2;                }                printf("2 * 3 = 6\n");                printf("6 * 4 = 24\n");                for(int i=1;i<=(n-4)/2+1;i++) printf("24 * 1 = 24\n");            }            else{                while(i>5){                    printf("%d - %d = %d\n",i,i-1,1);                    i-=2;                }                printf("2 + 4 = 6\n");                printf("3 * 6 = 18\n");                printf("18 + 5 = 23\n");                printf("23 + 1 = 24\n");                for(int i=1;i<=(n-5)/2;i++) printf("24 * 1 = 24\n");            }        }    }}

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