构造题,前面十几个手工处理....
n很大时有很多构造方法,一阵乱搞就可以了......
I Wanna Become A 24-Point Master
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 128 Accepted Submission(s): 36
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.
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 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
2015 Multi-University Training Contest 2
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <set>
#include <queue>
#include <stack>
#include <cmath>
#include <cstdlib>
using namespace std;
int dy[55];
void db(int x )
{
if ( x==1)
printf("-1\n");
if ( x==2)
printf("-1\n");
if ( x==3)
printf("-1\n");
if ( x==4)
{
puts("1 * 2");
puts("5 + 3");
puts("6 + 4");
}
if ( x==5)
{
puts("1 / 2");
puts("6 / 3");
puts("4 - 7");
puts("5 * 8");
}
if ( x==6)
{
puts("1 + 2");
puts("7 + 3");
puts("8 + 4");
puts("9 * 5");
puts("10 / 6");
}
if ( x==7)
{
puts("1 * 2"); // 8
puts("3 / 4"); // 9
puts("8 - 9"); // 10
puts("5 + 6"); // 11
puts("11 / 7"); // 12
puts("10 / 12"); // 13
}
if ( x==8)
{
puts("1 + 2"); // 9
puts("3 + 9"); // 10
puts("4 - 5"); // 11
puts("11 * 6"); // 12
puts("12 * 7"); // 13
puts("13 * 8"); // 14
puts("10 + 14"); // 15
}
if ( x==9 )
{
puts("1 + 2"); // 10
puts("10 + 3");
puts("11 + 4");
puts("12 + 5");
puts("13 + 6"); // 14
puts("14 / 7"); // 15
puts("15 + 8");
puts("16 + 9");
}
if ( x== 10)
{
puts("1 / 2"); //11
puts("3 / 4"); //12
puts("5 / 6"); // 13
puts("7 / 8"); // 14
puts("9 + 10"); //15
puts("15 + 11");
puts("16 + 12");
puts("17 + 13");
puts("18 + 14");
}
if ( x== 11)
{
puts("1 + 2"); //12
puts("12 / 3"); // 13
puts("13 * 4"); // 14
puts("14 / 5"); //15
puts("15 * 6"); //16
puts("16 / 7"); //17
puts("17 * 8"); // 18
puts("18 / 9"); //19
puts("19 + 10"); //20
puts("20 + 11"); // 21
}
if ( x== 12)
{
puts("1 + 2"); // 13
puts("13 + 3"); // 14
puts("14 - 4");
puts("15 + 5");
puts("16 - 6");
puts("17 + 7");
puts("18 - 8");
puts("19 + 9");
puts("20 - 10");
puts("21 + 11");
puts("22 - 12");
}
if (x == 24)
{
puts("1 - 2");
puts("25 * 3");
puts("26 * 4");
puts("27 * 5");
puts("28 * 6");
puts("29 * 7");
puts("30 * 8");
puts("31 * 9");
puts("32 * 10");
puts("33 * 11");
puts("34 * 12");
puts("35 * 13");
puts("36 * 14");
puts("37 * 15");
puts("38 * 16");
puts("39 * 17");
puts("40 * 18");
puts("41 * 19");
puts("42 * 20");
puts("43 * 21");
puts("44 * 22");
puts("45 * 23");
puts("46 + 24");
}
if (x == 26)
{
puts("1 / 2"); // 27
puts("3 / 4"); // 28
puts("27 + 28"); // 29
puts("5 - 29"); // 30
puts("30 * 6");
puts("31 / 7");
puts("32 * 8");
puts("33 / 9");
puts("34 * 10");
puts("35 / 11");
puts("36 * 12");
puts("37 / 13");
puts("38 * 14");
puts("39 / 15");
puts("40 * 16");
puts("41 / 17");
puts("42 * 18");
puts("43 / 19");
puts("44 * 20");
puts("45 / 21");
puts("46 * 22");
puts("47 / 23");
puts("24 - 25");
puts("49 * 26");
puts("48 + 50");
}
}
void solve25()
{
int n = 25;
int t = 1, e = dy[n + 1];
printf("%d + %d\n", dy[t], dy[t + 1]);
t += 2;
for(int i = 0; i < 22; i++)
{
printf("%d + %d\n", dy[t++], e++);
}
printf("%d / %d\n", e++, dy[t++]);
}
void xiao(int n, int c)
{
int t = 1, e = n + 1;
printf("%d - %d\n", t, t + 1);
t += 2;
for(int i = 2; i < c; i++)
{
printf("%d * %d\n", t++, e++);
}
printf("%d + %d\n", e++, t++);
int i;
//printf("n - c = %d\n", n - c);
for(i = 1; i < n - c; i++)
{
dy[i] = t++;
}
for(; i <= 55; i++)
{
dy[i] = e++;
}
//printf("*****%d\n", dy[26]);
//for(int i = 1; i <= 26; i++) printf("%d ", dy[i]);printf("\n");
}
void solve24(int n)
{
if(n == 26)
{
return;
}
for(int i = 1; i <= 88; i++) dy[i] = i;
if(n >= 27)
{
xiao(n, n - 25);
}
solve25();
}
void solve13(int n)
{
for(int i = 1; i < 55; i++) dy[i] = i;
if(n > 13)
{
xiao(n, 2 * n - 26);
}
int m = 26 - n;
int t = 1, e = dy[m + 1];
if(m > 3)
{
printf("%d + %d\n", dy[t], dy[t + 1]);
t += 2;
}
else
{
e--;
}
for(int i = 0; i < m - 4; i++)
{
printf("%d + %d\n", dy[t++], e++);
}
printf("%d / %d\n", e++, dy[t++]);
printf("%d + %d\n", dy[t++], e++);
}
void solve(int n)
{
if(n>=1&&n<=12) db(n);
else if(n==24||n==26) db(n);
else if(n > 24)
{
solve24(n);
}
else if(n >= 13)
{
solve13(n);
}
}
int main()
{
int n;
while(scanf("%d", &n) != EOF)
{
solve(n);
}
return 0;
}
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