24 Game CodeForces
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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 thema and b, erase them from the sequence, and append to the sequence eithera + b, or a - b, or a × b.
After n - 1 steps there is only one number left. Can you make this number equal to24?
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 followingn - 1 lines print the required operations an operation per line. Each operation should be in form: "aop b =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.
1
NO
8
YES8 * 7 = 566 * 5 = 303 - 4 = -11 - 2 = -130 - -1 = 3156 - 31 = 2525 + -1 = 24
<4肯定组不成24,
>=4,偶数的话最少4个数就可以出来24,奇数的话,5个数就可以,让剩下的数,后一个减去前一个,只保留一个1,然后不断地*24就好了
#include <cstdio> using namespace std; int main() { int n; while(~scanf("%d",&n)){ if(n<4){ printf("NO\n"); } else{ printf("YES\n"); if(n&1){ printf("3 + 4 = 7\n"); printf("7 + 5 = 12\n"); printf("12 * 2 = 24\n"); printf("24 * 1 = 24\n"); while(n>5){ printf("%d - %d =1\n",n,n-1); printf("24 * 1 =24\n"); n-=2; } } else{ printf("2 * 3 = 6\n"); printf("4 * 6 = 24\n"); printf("24 * 1 = 24\n"); while(n>4){ printf("%d - %d = 1\n",n,n-1); printf("24 * 1 = 24\n"); n-=2; } } } } }
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