CSU 1392 number Trick (number theory)

Test instructions: Enter a number x from small to large output all satisfied (k*leny+y) *x=y*10+k k*leny+y

Idea: Because X is a floating-point number, both sides of the equation are multiplied by 1e4.

Converts the equation to K (leny*x-1) *1e4=y (1E5-X*1E4)

It is possible to enumerate Len K to find out if there are any time and conditions for the presence of Y

It's a little bit more than a pit, but it's okay to ask Kay or die.

The first conversion of X should pay attention to the floating point error x= (int) (lx*10000+0.5);

Second, because the two sides of the equation may be 0 to find the Y is also 0 that is the number of single-digit

So we have a special sentence for this situation.

printf ("%d", k);
if (y>0) printf ("%d", y);
printf ("\ n");

#include <cstdio> #include <cstring> #include <iostream> #include <algorithm>using namespace    STD; #define LL long longint len (int n) {int res=1;    for (int i=1;i<n;i++) {res*=10; } return res;    int digit (long long num) {int cnt=0;        while (num) {num/=10;    cnt++; } return CNT;}    int main () {Long long leny,y;    int k;    Double LX;    int i,j;        while (scanf ("%lf", &AMP;LX)!=eof) {int ok=0;        ll x= (int) (lx*10000+0.5);            if (lx>=10) {printf ("No solution\n");        Continue        } ll Leny=1; for (i=0;i<=7;i++) {for (k=1;k<=9;k++) {long temp= (long Long) k* (lo                ng long) (leny*x) -1e4);                Long Long temp1=1e5-x;                if (temp%temp1!=0) continue;                Y=TEMP/TEMP1;                    if (digit (y) ==i) {if (ok==0) ok=1;      printf ("%d", k);              if (y>0) printf ("%d", y);                printf ("\ n");        }} leny*=10;    } if (ok==0) printf ("No solution\n"); } return 0;}

　　

CSU 1392 number Trick (number theory)