Alexandra and A*b problem
Time limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 802 Accepted Submission (s): 211
Problem Descriptionalexandra has a little brother. He is new to programming. One day he is solving the following problem:given the positive integers A and B, output a*b.
This problem was even easier than the last one. Alexandra can ' t wait to give him another task:given a positive integer A and a string s (s only contains numbers.), find T He minimum positive integer B, such that S is a substring of T, where T is the decimal notation of a*b.
See the sample for better understanding.
Note:s can contain leading zeros, but T can ' t.
Inputthere is multiple test cases (no more than 500). Each case contains a positive integer A and a string S.
A≤10,000,≤ |s| 8 .
Outputfor each case, output the required B. It is guaranteed this such B always exists.
to C + + programmers:if you want to output 64-bit integers and use "%i64d" specifier or cout.
Sample INPUT6 896 192 00861 1
Sample Output3250431 Test Instructions: Give a number A, and a string s (lens<=8) to find a minimum number B, so that s is a continuous sub-sequence of a*b = T. Problem Solving: Very ingenious topic. Suppose t = xsy, then we can write the expression T = (X*10^lens+s) *10^len+y, and because t%a = = 0 So for each part of the smallest T-equation, we can take a modulo, so we can know x<a, because if x >= A, we can make the x smaller by the modulo operation, then if s[0]==0, then the lower limit of X is 1, otherwise the lower limit of x is 0, then for 10^len we can also know 10^len<= a <= 10^4, and then we turn out: set k = (x* 10^lens+s) *10^len SO (k+y)%a=0y = (-k%a+a)%a (Y<10^len) So the final answer is obtained by enumerating the X,10^len.
#include <cstdio>#include<cstring>#include<vector>#include<algorithm>using namespaceStd;typedefLong LongLL;intMain () {LL A; Chars[ -]; while(SCANF ("%lld%s", &a,s)! =EOF) { intLen =strlen (s); if(strcmp (s),"0")==0) {printf ("0\n"); Continue; } LL LS=1, mid=0; for(intI=0; i<len;i++) ls*=Ten; for(intI=0; i<len;i++) {Mid= mid*Ten+s[i]-'0'; } LL ans= -1; for(LL i=1; i<=10000; i*=Ten){ for(LL j= (s[0]=='0'); j<a;j++) {LL T= (j*ls+mid) *i; LL y= (a-t%a)%A; if(y>=i)Continue; if(ans<0) ans = t+y; Elseans = min (t+Y,ans); }} printf ("%i64d\n", ans/a); } return 0;}
HDU 5109 (number of constructs + degree of understanding of modulo)
