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

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题意：有n张卡片，每张卡片的价格为c[i]，对于第i张卡片，可以从格子x跳到x+l[i]或者x-l[i]，问走完所有格子的最小费用为多少

对于卡片a、b，假设a卡片走x步， b卡片走y步，则一共走了 ax+by = c 步，当且仅当c是gcd(a, b)的整数倍时有解，所以最小步数为gcd（a,b）

所以选出的卡片的gcd应当等于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();}}

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