标签：codeforces 数学 构造
题目链接：http://codeforces.com/problemset/problem/468/A
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
思路：
答案有很多种，但最终目的都一样是构造出24点，显然小于4是不能构造成功的。对于大于等于4的数字，我们先分别打表出n等于4和5的时候的构造方法，然后再把大于5的数字全部转化（printf("%d - %d = 1\n",i,i-1);）为1，然后再printf("24 * 1 = 24\n");这样最后就只有24点保留了！
代码如下：
#include <cstdio>
using namespace std;
int main()
{
int n;
while(~scanf("%d",&n))
{
if (n < 4)
{
printf("NO");
continue;
}
printf("YES\n");
if (n&1)//奇数
{
printf("3 + 4 = 7\n");
printf("7 + 5 = 12\n");
printf("12 * 2 = 24\n");
printf("24 * 1 = 24\n");
}
else
{
printf("1 * 2 = 2\n");
printf("2 * 3 = 6\n");
printf("6 * 4 = 24\n");
}
for (int i = n; i > 5; i -= 2)//全部转化为1
{
printf("%d - %d = 1\n",i,i-1);
printf("24 * 1 = 24\n");
}
}
return 0;
}
CodeForces 468A. 24 Game（数学构造）
标签：codeforces 数学 构造
原文地址：http://blog.csdn.net/u012860063/article/details/39453807
