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
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.
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
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
#include <bits/stdc++.h>
using namespace std;
const int maxn = ;
const char str[][maxn] = {
"-1",
"-1",
"-1",
"-1",
"1 * 2\n5 + 3\n6 + 4",
"1 * 2\n6 * 3\n7 - 4\n8 / 5",
"1 + 2\n7 + 3\n8 + 4\n9 + 5\n10 - 6",
"1 + 2\n8 + 3\n4 + 5\n10 + 6\n11 / 7\n9 + 12",
"1 + 2\n9 + 3\n4 + 5\n11 - 6\n12 - 7\n13 / 8\n10 + 14",
"1 + 2\n10 + 3\n4 + 5\n12 + 6\n13 / 7\n11 - 14\n15 - 8\n16 + 9",
"1 + 2\n3 + 4\n12 + 5\n13 + 6\n14 / 7\n11 + 15\n8 - 9\n17 / 10\n16 + 18",
"1 + 2\n3 + 4\n13 / 5\n12 + 14\n15 - 6\n16 + 7\n17 - 8\n18 + 9\n19 - 10\n20 + 11",
"1 + 2\n3 + 13\n4 + 14\n5 + 6\n7 + 16\n8 + 17\n9 + 15\n10 + 19\n18 / 11\n20 / 12\n21 * 22",
"1 + 2\n3 + 4\n15 / 5\n14 - 16\n17 - 6\n18 + 7\n19 - 8\n20 + 9\n21 - 10\n22 + 11\n23 - 12\n24 + 13",
"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"
};
int main() {
int n;
while(~scanf("%d",&n)) {
if(n <= ) puts(str[n]);
else {
printf(str[],n+,n+,n+,n+,n+,n+,n+,n+,n+,n+);
int last = n + ;
printf("%d - %d\n",,);
for(int i = ; i <= n; ++i)
printf("%d * %d\n",i,last++);
printf("%d + %d\n",n + ,last);
}
}
return ;
}
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