将1-9填进算式中()()()()*()=()()()()()()()()*()=()()()() 九个数字不能重复使用,一个数字只能用一次( )( )( )( 3 ) 剩余1.5.6.7.8.9这六个数分别填进算式中,使算式成立*( 4 )-----------------( )( )( )( 2 )
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![将1-9填进算式中()()()()*()=()()()()()()()()*()=()()()() 九个数字不能重复使用,一个数字只能用一次( )( )( )( 3 ) 剩余1.5.6.7.8.9这六个数分别填进算式中,使算式成立*( 4 )-----------------( )( )( )( 2 )](/uploads/image/z/6967347-51-7.jpg?t=%E5%B0%861-9%E5%A1%AB%E8%BF%9B%E7%AE%97%E5%BC%8F%E4%B8%AD%28%29%28%29%28%29%28%29%2A%28%29%3D%28%29%28%29%28%29%28%29%28%29%28%29%28%29%28%29%2A%28%29%3D%28%29%28%29%28%29%28%29+%E4%B9%9D%E4%B8%AA%E6%95%B0%E5%AD%97%E4%B8%8D%E8%83%BD%E9%87%8D%E5%A4%8D%E4%BD%BF%E7%94%A8%2C%E4%B8%80%E4%B8%AA%E6%95%B0%E5%AD%97%E5%8F%AA%E8%83%BD%E7%94%A8%E4%B8%80%E6%AC%A1%28+%29%28+%29%28+%29%28+3+%29+%E5%89%A9%E4%BD%991.5.6.7.8.9%E8%BF%99%E5%85%AD%E4%B8%AA%E6%95%B0%E5%88%86%E5%88%AB%E5%A1%AB%E8%BF%9B%E7%AE%97%E5%BC%8F%E4%B8%AD%EF%BC%8C%E4%BD%BF%E7%AE%97%E5%BC%8F%E6%88%90%E7%AB%8B%2A%28+4+%29-----------------%28+%29%28+%29%28+%29%28+2+%29)
将1-9填进算式中()()()()*()=()()()()()()()()*()=()()()() 九个数字不能重复使用,一个数字只能用一次( )( )( )( 3 ) 剩余1.5.6.7.8.9这六个数分别填进算式中,使算式成立*( 4 )-----------------( )( )( )( 2 )
将1-9填进算式中()()()()*()=()()()()
()()()()*()=()()()() 九个数字不能重复使用,一个数字只能用一次
( )( )( )( 3 ) 剩余1.5.6.7.8.9这六个数分别填进算式中,使算式成立
*( 4 )
-----------------
( )( )( )( 2 )
将1-9填进算式中()()()()*()=()()()()()()()()*()=()()()() 九个数字不能重复使用,一个数字只能用一次( )( )( )( 3 ) 剩余1.5.6.7.8.9这六个数分别填进算式中,使算式成立*( 4 )-----------------( )( )( )( 2 )
haha...有意思.答案居然是唯一的.
1963*4 = 7852
#include
#include
#include
int main(void)
{
int i,k;
char str1[5];
char str2[5];
int n[10];
for (k=0; k<999; k++)
{
i = k*10+3;
int res = i*4;
if (res >= 10000)
break;
if (res-res/10*10 == 2)
{
sprintf(str1, "%04d", i);
sprintf(str2, "%04d", res);
for (int j=1; j<10; j++)
n[j] = 0;
n[4] = 1;
for (int j=0; j
for (int j=0; j
bool flag = true;
for (int j=1; j<10; j++)
if (n[j] != 1)
{
flag = false;
break;
}
if (flag)
printf("%04d %04d success\n", i, res);
// else
// printf("%04d %04d fail\n", i, res);
}
}
system("PAUSE");
return 0;
}
1963*4=7852
n那个*是乘吗?