http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1392

1392: Number Trick

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 200  Solved: 36
[Submit][Status][Web Board]

Description

Input

Output

Sample Input

2.6

Sample Output

135

270

135135

270270

HINT

分析：

给一个小数X，找个A使得：AX=（A循环左移一位）

首先，假设A为一个满足题目条件的数，有n个数位，且最高位数字为A 0 。

那么可列出方程：AX=（A-A 0* 10 n-1 ）*10+A 0————>> A 0 *（10 n -1）=A*（10-X）

也就是说，只要n和A 0 确定了，A也就可以唯一确定。

那么我们只要直接枚举：n=(1->8) A 0 =(1->9)

计算出A是不是满足这两个条件就可以得出正确答案了。

AC代码：

 #include <cstdio>

 typedef long long ll;

 using namespace std;

 int power[],a[],ans=;

 float x,tmpx;//不知道为什么用double会wa

 ll f1,f2;

 int digit(ll tmp)

 {

   int tot=;

   while(tmp) tot++,tmp/=;

   return tot;

 }

 int head(ll tmp)

 {

   for (; tmp>=; tmp/=);

   return tmp;

 }

 int main()

 {

   power[]=;

   for (int i=; i<=; i++) power[i]=power[i-]*;

   scanf("%f",&x);

   if (x>=)

   {

     printf("No solution\n");

     return ;

   }

   for (int i=; i<=; i++)//枚举总共几个数位

     for (int j=; j<=; j++)//枚举开头数字

     {

       tmpx=-*x;

       f1=(int)tmpx;

       f2=(ll)*(power[i]-)*j;

       if (f2%f1!=) continue;

       f2/=f1;

       if (digit(f2)==i && head(f2)==j) a[++ans]=f2;

     }

   if (ans==) printf("No solution\n");

     else for (int i=; i<=ans; i++) printf("%d\n",a[i]);

   return ;

 }

学长打表：

 //#include <cstdio>

 //#include <cstring>

 //

 //using namespace std ;

 //

 //typedef long long LL ;

 //

 //LL change( LL a ) {

 //    LL data[32] ;

 //    int cur = 0 ;

 //    while( a ) {

 //        data[cur++] = a % 10LL ;

 //        a /= 10LL ;

 //    }

 //    LL ans = 0 ;

 //    for( int i = cur - 2 ; i >= 0 ; i -- ) {

 //        ans = ans * 10 + data[i] ;

 //    }

 //    ans = ans * 10 + data[cur-1] ;

 //    return ans ;

 //}

 //

 //int main() {

 //

 //    freopen("out.txt","w",stdout);

 //    int cnt = 0 ;

 //    printf("void init() {\n") ;

 //    for( LL a = 1LL ; a <= 100000000LL ; a ++ ) {

 //        LL b = change( a ) ;

 //        if( 10000LL * b % a == 0 ) {

 //            LL t = 10000LL * b / a ;

 //            //ans[t].push_back(a) ;

 //            printf("ans[%lld].push_back(%lld);" , t , a ) ;

 //            if( ++cnt == 100 ) {

 //                printf("\n") ;

 //                cnt = 0 ;

 //            }

 //        }

 //    }

 //    printf("\n}\n") ;

 //

 //    return 0 ;

 //}

 #include <vector>

 #include <cstdio>

 #include <cstring>

 using namespace std ;

 typedef long long LL ;

 vector<int> ans[+] ;

 void init() {

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 }

 int main() {

     init() ;

     double n ;

     while( scanf("%lf" , &n ) ==  ) {

         int a = (int)( n *  + 0.5 ) ;

         if( ans[a].size() ==  ) printf("No solution\n") ;

         else {

             for( int i =  ; i < ans[a].size() ; i ++ ) {

                 printf("%d\n" , ans[a][i] ) ;

             }

         }

     }

     return  ;

 }

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