[HDOJ5339] Untitled

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=5339

Give you n number b1...bn and a number A, let you take a to the b1...bn inside the number of modulo, the modulus after the value of 0 of the minimum number. You can repeat the value in B.

The first thing to be clear is that it doesn't make any sense to take a number to a smaller number and then take the modulus of a larger number. Therefore, it is meaningless to repeat the values, so the array B is sorted from large to small.

After the state is compressed, enumerate the numbers in B. You can also DFS out the results.

1#include <cstdio>2#include <cstdlib>3#include <cstring>4#include <algorithm>5#include <iostream>6#include <cmath>7#include <queue>8#include <map>9#include <stack>Ten#include <list> One#include <vector> A  - using namespacestd; -  the Const intMAXN =100010; - Const intINF =1<< -; - intN, A; - intB[MAXN]; +  -InlineintMinintAintb) { +     returnA < b?a:b; A } at intMain () { -     //freopen ("in", "R", stdin); -     intT; -scanf"%d", &T); -      while(t--) { -         intAns =INF; inscanf"%d%d", &n, &a); -          for(inti =0; I < n; i++) { toscanf"%d", &b[i]); +         } -Sort (b, b+n, greater<int>()); the         ints = (1<<n); *         intCNT; $          for(inti =1; I < S; i++) {Panax Notoginseng             intTMP =A; -CNT =0; the              for(intj =0; J < N; J + +) { +                 if(I & (1<<j)) { ATMP%=B[j]; thecnt++; +                 } -             } $             if(TMP = =0) { $Ans =min (ans, cnt); -             } -         } the         if(ans = =INF) { -printf"-1\n");Wuyi         } the         Else { -printf"%d\n", ans); Wu         } -     } About}

[HDOJ5339] Untitled