2015多校联合第二场5308I Wanna Become A 24-Point Master
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 i th 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 i th 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
Author
XJZX
Source
2015 Multi-University Training Contest 2
好歹这个题看明白问啥了￣□￣｜｜ 一丢丢思路都没有啊啊啊啊啊啊 度娘告诉我 要构造要手动拆 好吧你赢了 小数可以手动拆 大数呢 又说可以构造24 how?
(4*n/n)*(6*n/n)=24 当然了 凑2*12 或者3*8也没人拦着你 再找两个数减得0与剩下的数乘起来 所以 小数 3以下的凑不出来14一下的手动算 其实没多难 就是想不到T^T
#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
int main()
{
int n,tmp;
while(~scanf("%d",&n))
{
// printf("%d:\n",n);
if(n<=3)
printf("-1\n");
else if(n==4)
printf("1 * 2\n5 + 3\n6 + 4\n");
else if(n==5)
printf("1 * 2\n3 / 6\n4 - 7\n5 * 8\n");
else if(n==6)
printf("1 + 2\n3 + 4\n5 - 6\n7 + 8\n10 - 9\n");
else if(n==7)
printf("1 + 2\n3 + 8\n9 / 4\n10 + 5\n11 + 6\n12 + 7\n");
else if(n==8)
printf("1 + 2\n3 + 9\n4 - 5\n11 * 6\n12 * 7\n13 * 8\n10 + 14\n");
else if(n==9)
printf("1 + 2\n3 + 10\n4 / 5\n6 / 7\n8 / 9\n11 - 12\n15 - 13\n 16 - 14\n");
else if(n==10)
printf("1 + 2\n3 / 4\n5 / 6\n7 / 8\n9 / 10\n11 + 12\n16 + 13\n17 + 14\n18 + 15\n");
else if(n==11)
printf("1 + 2\n3 / 4\n5 / 6\n7 - 8\n15 * 9\n16 * 10\n17 * 11\n12 + 13\n19 + 14\n20 + 18\n");
else if(n==12)
printf("1 + 2\n3 - 4\n5 * 14\n6 * 15\n7 * 16\n8 * 17\n9 * 18\n10 * 19\n11 * 20\n12 * 21\n13 + 22\n");
else if(n==13)
printf("1 + 2\n3 / 4\n5 / 6\n7 - 8\n17 * 9\n18 * 10\n19 * 11\n20 * 12\n21 * 13\n22 + 14\n23 - 15\n24 - 16\n");
else
{
printf("1 + 2\n3 + 4\n5 + 6\n7 + 8\n9 + 10\n");
printf("%d + %d\n%d + %d\n%d + %d\n",n+1,n+2,n+3,n+4,n+5,n+6);
printf("%d / 11\n%d / 12\n",n+7,n+8);
printf("%d * %d\n",n+9,n+10);
printf("13 - 14\n");
tmp=n-14;
int i;
for(i=0;i<tmp;i++)
{
printf("%d * %d\n",n+12+i,15+i);
}
printf("%d + %d\n",n+11,n+12+tmp);
}
// printf("\n");
}
return 0;
}