HDU 1573 X问题 数论

思路：令LCM=lcm(a1,a2…an)，符合条件的第一个数一定≤LCM，后面的每一个数都是在前面的基础上加上LCM。

坑点：可能找不出符合条件的第一个数，这种情况要特判。

http://acm.hdu.edu.cn/showproblem.php?pid=1573

/*********************************************    Problem : HDU 1573    Author  : NMfloat    InkTime (c) NM . All Rights Reserved .********************************************/#include <map>#include <set>#include <queue>#include <stack>#include <cmath>#include <ctime>#include <cstdio>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>#define rep(i,a,b)  for(int i = (a) ; i <= (b) ; i ++)#define rrep(i,a,b) for(int i = (b) ; i >= (a) ; i --)#define repE(p,u) for(Edge * p = G[u].first ; p ; p = p -> next)#define cls(a,x)   memset(a,x,sizeof(a))#define eps 1e-8using namespace std;const int MOD = 1e9+7;const int INF = 0x3f3f3f3f;const int MAXN = 1e5+5;const int MAXE = 2e5+5;typedef long long LL;typedef unsigned long long ULL;int T,n,m,k;int fx[] = {0,1,-1,0,0};int fy[] = {0,0,0,-1,1};int a[15],b[15];int N;int gcd(int a1,int a2) {    int tmp;    while(a2) {        tmp = a1 % a2;        a1 = a2;        a2 = tmp;    }    return a1;}int lcm(int a1,int a2) {    return a1 * a2 / gcd(a1,a2);}void input() {    scanf("%d %d",&N,&m);    rep(i,1,m) scanf("%d",&a[i]);    rep(i,1,m) scanf("%d",&b[i]);}void solve() {    int lcm_num = 1;    rep(i,1,m) {        lcm_num = lcm(lcm_num,a[i]);    }    int pos = 0;    rep(i,1,lcm_num) {        int ok = 0;        rep(j,1,m) {            if(i % a[j] != b[j]) {ok = 1; break;}         }        if(ok == 0) {pos = i ; break;}     }    if(pos == 0) { puts("0"); return ;}    int ans = 1;    if(N < pos) ans = 0;    else {        N -= pos;        ans += N / lcm_num;            }    printf("%d\n",ans);}int main(void) {    //freopen("a.in","r",stdin);    scanf("%d",&T); while(T--) {    //while(~scanf("%d %d",&n,&m)) {    //while(scanf("%d",&n),n) {        input();        solve();    }    return 0;}