Codeforces Round #290 (Div. 1) C Fox And Jumping

題意：有n張卡片，每張卡片的價格爲c[i]，對於第i張卡片，可以從格子x跳到x+l[i]或者x-l[i]，問走完所有格子的最小費用爲多少

對於卡片a、b，假設a卡片走x步， b卡片走y步，則一共走了 ax+by = c 步，當且僅當c是gcd(a, b)的整數倍時有解，所以最小步數爲gcd（a,b）

所以選出的卡片的gcd應當等於1

用map搞下就可以了

#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<stdio.h>
#include<algorithm>
#include<cmath>
#include<set>
#include<map>
#include<queue>
#include<vector>
using namespace std;


#define inf 0x3f3f3f3f
#define eps 1e-9
#define mod 1000000007
#define FOR(i,s,t) for(int i = s; i < t; ++i )
#define REP(i,s,t) for( int i = s; i <= t; ++i )
#define LL long long
#define ULL unsigned long long
#define pii pair<int,int>
#define MP make_pair
#define lson id << 1 , l , m
#define rson id << 1 | 1 , m + 1 , r 
#define maxn ( 500 + 10 )
#define maxe ( 500+10 )

map< int, int > gao;
map< int , int >:: iterator it;
int l[maxn];
int c[maxn];

int gcd( int a, int b ) {
	return b == 0 ? a : gcd( b, a%b );
}

int main () {
	int n;
	while( ~scanf("%d", &n ) ) {
		for( int i = 0; i < n; ++i )
			scanf("%d", &l[i] );
		for( int i = 0; i < n; ++i )
			scanf("%d", &c[i] );
		gao[0] = 0;
		for( int i = 0; i < n; ++i ) {
			for( it = gao.begin(); it != gao.end(); ++it ) {
				int t1 = it->first, t2 = it->second;
				int tmp = gcd( t1, l[i] );
				if( gao.count( tmp ) )
					gao[tmp] = min( gao[tmp], t2 + c[i] );
				else gao[tmp] = t2 + c[i];
			}
		}
		if( !gao.count( 1 ) )
			printf( "-1\n" );
		else printf( "%d\n", gao[1] );
		gao.clear();
	}
}