I Wanna Become A 24-Point Master
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 60 Accepted Submission(s): 16
Special 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 with n numbers,
which are all equal to n.
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 with n numbers,
which are all equal to n.
Input
There are no more then 100 testcases and there are no more then 5 testcases with n≥100.
Each testcase contains only one integer n (1≤n≤105)
Each testcase contains only one integer n (1≤n≤105)
Output
For each testcase:
If there is not any way to get 24 points, print a single line with -1.
Otherwise, let A be
an array with 2n−1 numbers
and at firsrt Ai=n (1≤i≤n).
You need to print n−1 lines
and the ith
line contains one integer a,
one char b and
then one integer c, where 1≤a,c<n+i and b is
"+","-","*" or "/". This line means that you let Aa and Ac do
the operation b and
store the answer into An+i.
If your answer satisfies the following rule, we think your answer is right:
1. A2n−1=24
2. Each position of the array A is
used at most one tine.
3. The absolute value of the numerator and denominator of each element in array A is
no more than 109
If there is not any way to get 24 points, print a single line with -1.
Otherwise, let A be
an array with 2n−1 numbers
and at firsrt Ai=n (1≤i≤n).
You need to print n−1 lines
and the ith
line contains one integer a,
one char b and
then one integer c, where 1≤a,c<n+i and b is
"+","-","*" or "/". This line means that you let Aa and Ac do
the operation b and
store the answer into An+i.
If your answer satisfies the following rule, we think your answer is right:
1. A2n−1=24
2. Each position of the array A is
used at most one tine.
3. The absolute value of the numerator and denominator of each element in array A is
no more than 109
Sample Input
4
Sample Output
1 * 2
5 + 3
6 + 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能够用分数表示，不能截断小数。
分析：前几个数字打表吧。非常easy写的。可是我队员（GMY）也是写的非常辛苦啊。========(如今看到是改动过后的，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降低用于构造24的N的使用量。
且令b = |24-a*N| （a*N+b = 24 或者a*N-b = 24)构造这个等式就能够了。
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就可以。
显然left >= 2 （7特判了，就不怕有bug了）
假设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一定能够构造出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。
由于保证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24
int 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; //算出须要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0
printf("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 / N
if(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; }
//推断是加法还是减法
if(fx > 0)
printf("%d + %d\n",p,xp);
else
printf("%d - %d\n",p,xp);
}
}
}
return 0;
}