hdu5308 I Wanna Become A 24-Point Master(构造)

题目：http://acm.hdu.edu.cn/showproblem.php?pid=5308

题意：给定N个N,求出将这N个N构造成24的方法。

分析：要是想到用尽量少的N构造出24点（(N+N)/N * (N+N)/N * (N+N)/N * (N*N+N)/N），然后把多的N构造成0就好办了（(N-N)*N*.....）。

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;

void _15(int n)
{
	printf("1 + 2\n");//n+1
	printf("4 + 5\n");//n+2
	printf("7 + 8\n");//n+3
	printf("10 + 11\n");//n+4
	printf("%d + 12\n",n+4);//n+5
	printf("%d / 3\n",n+1); // n+6 2
	printf("%d / 6\n",n+2);//n+7   2
	printf("%d / 9\n",n+3);//n+8   2
	printf("%d / 13\n",n+5);//n+9  3
	printf("14 - 15\n");//n+10     0
	int c=n+10;
	for(int i=16;i<=n;i++)
	{
		printf("%d * %d\n",c,i);
		c++;
	}
	printf("%d * %d\n",n+6,n+7); //c+1
	printf("%d * %d\n",n+8,c+1); //c+1
	printf("%d * %d\n",c+2,n+9);
	printf("%d + %d\n",c+3,c);
}
void _14()
{
	int s1[100]={1,15,16,17,18,7,9,11,20,21,22,25,26};
	int s2[100]={2,3,4,5,6,8,10,12,13,14,19,23,24};
	char op[100]="-****+++//+--";
	for(int i=0;op[i];i++)
		printf("%d %c %d\n",s1[i],op[i],s2[i]);
}
void _13()
{
	int s1[100]={1,4,7,10,17,14,15,16,18,19,21,23};
	int s2[100]={2,5,8,11,12,3,6,9,13,20,22,24};
	char op[100]="+++++***";
	for(int i=0;op[i];i++)
		printf("%d %c %d\n",s1[i],op[i],s2[i]);
}

void _12()
{
	int s1[100]={1,3,5,7,9,11,13,15,17,19,21};
	int s2[100]={2,4,6,8,10,12,14,16,18,20,22};
	char op[100]="-----++++++";
	for(int i=0;op[i];i++)
		printf("%d %c %d\n",s1[i],op[i],s2[i]);
}
void _11()
{
	int s1[100]={1,12,13,14,15,7,17,9,18,19};
	int s2[100]={2,3,4,5,6,8,16,10,11,20};
	char op[100]="-****+++/+";
	for(int i=0;op[i];i++)
		printf("%d %c %d\n",s1[i],op[i],s2[i]);
}

void _10()
{
	int s1[100]={1,11,4,6,8,13,14,15,17};
	int s2[100]={2,3,5,7,9,10,12,16,18};
	char op[100]="-++++//++";
	for(int i=0;op[i];i++)
		printf("%d %c %d\n",s1[i],op[i],s2[i]);
}
void _9()
{
	int s1[100]={1,3,4,11,6,8,15,13};
	int s2[100]={2,10,5,12,7,14,9,16};
	char op[100]="++-+++/-";
	for(int i=0;op[i];i++)
		printf("%d %c %d\n",s1[i],op[i],s2[i]);
}
void _8()
{
	int s1[100]={1,9,10,11,12,13,14};
	int s2[100]={2,3,4,5,6,7,8};
	char op[100]="-***+++";
	for(int i=0;op[i];i++)
		printf("%d %c %d\n",s1[i],op[i],s2[i]);
}
void _7()
{
	int s1[100]={1,3,4,6,11,12};
	int s2[100]={2,8,5,10,7,9};
	char op[100]="++++/+";
	for(int i=0;op[i];i++)
		printf("%d %c %d\n",s1[i],op[i],s2[i]);
}
void _6()
{
	int s1[100]={1,3,7,5,9};
	int s2[100]={2,4,8,6,10};
	char op[100]="+++-+";
	for(int i=0;op[i];i++)
		printf("%d %c %d\n",s1[i],op[i],s2[i]);
}
void _5()
{
	printf("1 * 2\n");
	printf("6 * 3\n");
	printf("7 - 4\n");
	printf("8 / 5\n");
}
void _4()
{
	printf("1 * 2\n");
	printf("5 + 3\n");
	printf("6 + 4\n");
}

int main()
{
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		if(n>=15)
			_15(n);
		else if(n==14)
			_14();
		else if(n==13)
			_13();
		else if(n==12)
			_12();
		else if(n==11)
			_11();
		else if(n==10)
			_10();
		else if(n==9)
			_9();
		else if(n==8)
			_8();
		else if(n==7)
			_7();
		else if(n==6)
			_6();
		else if(n==5)
			_5();
		else if(n==4)
			_4();
		else
			printf("-1\n");
	}
	return 0;
}