CodeForces1214A

CodeForces1214A
说起来你们可能不信,这题硬生生卡了我(1h),我想了背包,扩欧,二分....等等一坨办法.结果最后还是用了(bfs)过的.
呃,代码一目了然,不用说吧...
(Code:)

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <string>
#include <vector>
#include <queue>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
#define MEM(x,y) memset ( x , y , sizeof ( x ) )
#define rep(i,a,b) for (int i = a ; i <= b ; ++ i)
#define per(i,a,b) for (int i = a ; i >= b ; -- i)
#define pii pair < int , int >
#define X first
#define Y second
#define rint read<int>
#define pb push_back
 
using std::set ;
using std::pair ;
using std::max ;
using std::min ;
using std::priority_queue ;
using std::vector ;
using std::swap ;
using std::sort ;
using std::unique ;
using std::greater ;
using std::queue ;
 
template < class T >
    inline T read () {
        T x = 0 , f = 1 ; char ch = getchar () ;
        while ( ch < '0' || ch > '9' ) {
            if ( ch == '-' ) f = - 1 ;
            ch = getchar () ;
        }
        while ( ch >= '0' && ch <= '9' ) {
            x = ( x << 3 ) + ( x << 1 ) + ( ch - 48 ) ;
            ch = getchar () ;
       }
   return f * x ;
}
 
template < class T >
    inline void write(T x) {
        static T stk[100], top = 0;
        if (x == 0) { putchar('0'); return; }
        if (x < 0) { x = -x; putchar('-'); }
        while (x) { stk[++top] = x % 10; x /= 10; }
        while (top) { putchar(stk[top--] + '0'); }
        putchar ( 10 ) ;
    }
 
const int N = 1e8 + 100 ;
 
int dollar , euro , tot ;
int d[] = { 1 , 2 , 5 , 10 , 20 , 50 , 100 } ;
int e[] = { 5 , 10 , 20 , 50 , 100 , 200 } ;
bool mk[N] ; queue < int > q ; 

inline void bfs (int x) {
    q.push ( x ) ; mk[x] = true ;
    while ( ! q.empty () ) {
        int j = q.front () ; q.pop () ;
        rep ( i , 0 , 6 ) {
            int tmp = j - d[i] ;
            if ( tmp >= 0 && ! mk[tmp] ) {
                mk[tmp] = true ;
                q.push ( tmp ) ;
            }
        }
        rep ( i , 0 , 5 ) {
            int tmp = j - e[i] ;
            if ( tmp >= 0 && ! mk[tmp] ) {
                mk[tmp] = true ;
                q.push ( tmp ) ;
            }
        }
    }
    return ;
}

signed main (int argc , char * argv[] ) {
    tot = rint () ; dollar = rint () ; euro = rint () ; MEM ( mk , 0 ) ;
    rep ( i , 0 , 6 ) d[i] *= dollar ; rep ( i , 0 , 5 ) e[i] *= euro ;
    bfs ( tot ) ; rep ( i , 0 , tot ) if ( mk[i] ) { write ( i ) ; break ; }
    return 0 ;
}