I Wanna Become A 24-Point MasterTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 60Accepted Submission(s): 16Special Judge
Problem Description
Recently Rikka falls in love with an old but interesting game — 24 points. She wants to become a master of this game, so she asks Yuta to give her some problems to practice.Quickly, Rikka solved almost all of the problems but the remained one is really difficult:In this problem, you need to write a program which can get 24 points withnumbers, which are all equal to.
Input
There are no more then 100 testcases and there are no more then 5 testcases with. Each testcase contains only one integer
Output
For each testcase:If there is not any way to get 24 points, print a single line with -1.Otherwise, letbe an array withnumbers and at firsrt. You need to printlines and theth line contains one integer, one charand then one integer c, whereis "+","-","*" or "/". This line means that you letdo the operationand store the answer into.If your answer satisfies the following rule, we think your answer is right:1.2. Each position of the arrayis used at most one tine.3. The absolute value of the numerator and denominator of each element in arrayis no more than
Sample Input
4
Sample Output
1 * 25 + 36 + 4
Source
题意：
输入N。用N个N通过加减乘除得到24.中间过程得到的数字可以用分数表示，不能截断小数。
分析：前几个数字打表吧。很容易写的。但是我队员（GMY）也是写的很辛苦啊。========(现在看到是修改过后的，所以打表的部分少了）
后面的数字这样处理：
选a个N相加 使得|a*N-24|最小,这样做的目的是为了减少用于构造24的N的使用量。
且令b = |24-a*N| （a*N+b = 24 或者a*N-b = 24)构造这个等式就可以了。
那么用b+1个N就能构造 b了 。方法： （N *b)/N = b
剩下left = N – a – b – 1个N。用left个N构造0即可。
显然left >= 2 （7特判了，就不怕有bug了）
如果left 为奇数
构造 （N*(left/2) – N(left/2))/N = 0
否则 构造 N*(left/2) – N(left/2) = 0
所以left个N一定可以构造出0的
最后看b > 0 那么用a*N – b
否则用a*N + b
当然要特别考虑b = 0,的情况。
因为保证left > 0 了。所以先构造出0.方便后面的运算。
#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int main(){int n;while(~scanf("%d",&n)){if( n == 1 || n == 2 || n == 3 ){printf("-1\n");}else if( n == 4 )printf("1 * 2\n5 + 3\n6 + 4\n");else if( n == 5 )printf("1 * 2\n6 * 3\n7 – 4\n8 / 5\n");else if( n == 7 )printf("1 + 2\n8 + 3\n4 + 5\n10 + 6\n11 / 7\n9 + 12\n"); //为了保证left > 1 n = 7 特判else {int i,j,k,q,p,left,need;for(i = 0;;i++){if(abs(i*n – 24) <= abs(i*n+n-24) )break;}int a = i; //算出几个N相加会最接近于24int fx = 24 – a*n; //算出 24 – a * N 关于0的大小if( abs(24-a*n) > 0 )need = abs(24-a*n) + 1;else need = 0;//算出需要need个N来构造 出|24 – a*N|left = n – need – i; //用left个N构造出0//用left个N构造0printf("1 – 2\n");p = n + 1;i = 3;j = 0;for(;i <= (left – left% 2); ){if(j & 1 ){printf("%d – %d\n",p++,i++);}else printf("%d + %d\n",p++,i++);j ^= 1;}//left 为奇数，多计算一次 0 / Nif(left % 2 == 1){printf("%d / %d\n",p++,i++);}//构造a个N相加的和q = i;for(i = 0;i < a; i++){printf("%d + %d\n",p++,q++);}int xp;if(need > 0){//need = 2 直接做除法，，否则先做加法，最后做除法if(need != 2){printf("%d + %d\n",q,q+1);q += 2;xp = p + 1;for(; q < n; ){printf("%d + %d\n",xp++,q++);}printf("%d / %d\n",xp++,q++);}else {printf("%d / %d\n",q,q+i);xp = p + 1;}//判断是加法还是减法if(fx > 0)printf("%d + %d\n",p,xp);elseprintf("%d – %d\n",p,xp);}}}return 0;}
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有时我们选择改变，并非经过深思熟虑，而更像是听见了天地间冥冥中的呼唤，
