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Subject: [JSJ] kadai2 kippu
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Date: Fri, 14 May 1999 04:41:31 +0900
From: Toshiyuki Kurosaki <g94p0407@mse.waseda.ac.jp>
Status: ORS


とちゅうまでしか出来ていませんが、とりあえず提出します。
今日もう少しかんがえて提出します。

#include <stdio.h>
#include <string.h>

void kippu2number();
void sum2ten(int k);
void back(int p);
void copy(int q);
void can();
void combi_sum2ten(int i);

int a[4];
int p[4];
double number[4] , t[4][4] ;
int kippu;

void main()
{
	int i;
	kippu2number();

	for( i= 0 ; i <=3 ; i++)	p[i] = i;
	
	combi_sum2ten(0);

}

/* 数字を入力してそれをa[]に代入する関数 */

void kippu2number()
{
	
	printf("４桁の数字を入力してください。#");
	scanf("%d",&kippu);
	a[0] = kippu / 1000;
	a[1] = (kippu - a[0] * 1000) / 100;
	a[2] = (kippu - a[0] * 1000 - a[1] * 100) / 10;
	a[3] = kippu - a[0] * 1000 - a[1] * 100 - a[2] * 10;
	printf("%d ,%d ,%d ,%d\n",a[0],a[1],a[2],a[3]);

}

/* k個の数字で10を作る */

void sum2ten(int k)
{
	
	if(k == 2){
		if( number[0] + number[1] == 10) can();
		else if( number[0] - number[1] == 10) can();
		else if( number[1] - number[0] == 10) can();
		else if( number[0] * number[1] == 10) can();
		else if( number[0] / number[1] == 10) can();
		else if( number[1] / number[0] == 10) can();
		else printf("can't\n");
	}

	if(k != 2){

		number[k-2] = number[k-2] + number[k-1];
		copy(k-1);
		sum2ten(k-1);
		back(k-1);
		
		number[k-1] = number[k-1] - number[k-1];
		copy(k-1);
		sum2ten(k-1);
		back(k-1);
		
		number[k-2] = number[k-2] - number[k-1];
		copy(k-1);
		sum2ten(k-1);
		back(k-1);
		
		number[k-2] = number[k-2] * number[k-1];
		copy(k-1);
		sum2ten(k-1);
		back(k-1);

		number[k-2] = number[k-2] / number[k-1];
		copy(k-1);
		sum2ten(k-1);
		back(k-1);

		number[k-2] = number[k-1] / number[k-2];
		copy(k-1);
		sum2ten(k-1);
		back(k-1);
	}
}

void copy(int q)
{
	int i;
	for(i=0 ; i<4 ; i++)	t[q][i] = number[i];
}

void back(int p)
{
	int i;
	for(i=0 ; 1<4 ; i++)	number[i] = t[p][i];
}

void can()
{
	printf("can\n");
}

void combi_sum2ten(int i)
{
	int j,t;

	if(i < 3){
		for( j = i ; j<=3 ; j++){
			t = p[i];
			p[i] = p[j];
			p[j] = t;
			combi_sum2ten(i+1);
			t = p[i] ;
			p[i] = p[j];
			p[j] = t;
		}
	}
	else{
		for(j=0;j<=3;j++){
			number[j] = a[p[j]];
		}
	sum2ten(4);
	}
}