2015 Multi-University Training Contest 2 hdu 5308 I Wanna Become A 24-Point Master
I Wanna Become A 24-Point Master
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 897 Accepted Submission(s): 379
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)
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 band 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 band 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
解题：打表+规律
可以发现当n等于12时，可以求解由
至于其余的数字，假设我们取n = 14 由于得到24，前面n个我们只用到了12个，那么我们可以将13 - 14，然后再加上 24 仍然是24
如果是15 ，我们可以13 - 14，然后差乘以 15 最后积加上24.。。以此类推
$\frac{n + n + n + n}{n} \times \frac{n + n + n + n + n + n}{n} = 24$
妈拉个巴子，latex公式不管用了
好吧
(n+n+n+n)/n = 4;
(n+n+n+n+n+n)/n = 6;
4*6 = 24
所以当n大于12的时候，已经很明显可以解决了
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int maxn = 2100; 4 const char str[16][maxn] = { 5 "-1", 6 "-1", 7 "-1", 8 "-1", 9 "1 * 2\n5 + 3\n6 + 4", 10 "1 * 2\n6 * 3\n7 - 4\n8 / 5", 11 "1 + 2\n7 + 3\n8 + 4\n9 + 5\n10 - 6", 12 "1 + 2\n8 + 3\n4 + 5\n10 + 6\n11 / 7\n9 + 12", 13 "1 + 2\n9 + 3\n4 + 5\n11 - 6\n12 - 7\n13 / 8\n10 + 14", 14 "1 + 2\n10 + 3\n4 + 5\n12 + 6\n13 / 7\n11 - 14\n15 - 8\n16 + 9", 15 "1 + 2\n3 + 4\n12 + 5\n13 + 6\n14 / 7\n11 + 15\n8 - 9\n17 / 10\n16 + 18", 16 "1 + 2\n3 + 4\n13 / 5\n12 + 14\n15 - 6\n16 + 7\n17 - 8\n18 + 9\n19 - 10\n20 + 11", 17 "1 + 2\n3 + 13\n4 + 14\n5 + 6\n7 + 16\n8 + 17\n9 + 15\n10 + 19\n18 / 11\n20 / 12\n21 * 22", 18 "1 + 2\n3 + 4\n15 / 5\n14 - 16\n17 - 6\n18 + 7\n19 - 8\n20 + 9\n21 - 10\n22 + 11\n23 - 12\n24 + 13", 19 "1 + 2\n3 + %d\n4 + %d\n5 + 6\n7 + %d\n8 + %d\n9 + %d\n10 + %d\n%d / 11\n%d / 12\n%d * %d\n" 20 }; 21 int main() { 22 int n; 23 while(~scanf("%d",&n)) { 24 if(n <= 13) puts(str[n]); 25 else { 26 printf(str[14],n+1,n+2,n+4,n+5,n+3,n+7,n+6,n+8,n+9,n+10); 27 int last = n + 12; 28 printf("%d - %d\n",13,14); 29 for(int i = 15; i <= n; ++i) 30 printf("%d * %d\n",i,last++); 31 printf("%d + %d\n",n + 11,last); 32 } 33 } 34 return 0; 35 }