C. Alarm Clocks Everywhere
思路:题意是给你n个任务,你可以任意选择一个开始时间,并从给定序列中选择一个间隔,使得能完成这n个任务。
就是求出来n-1个差,然后求出来最大公约数,然后判断这m个间隔是否存在一个是这个数的因子即可。
#include<cstdio>
#include<cmath>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1|1
const int inf = 0x3f3f3f3f;
const int maxn = 3e5 + 10;
ll a[maxn], p[maxn];
ll d[maxn];
ll gcd(ll x, ll y)
{
return y == 0 ? x : gcd(y, x % y);
}
int main()
{
int n, m;
scanf("%d%d", &n, &m);
for(int i = 1; i <= n; ++i)
{
scanf("%I64d", &a[i]);
}
for(int i = 2; i <= n; ++i)
{
d[i] = a[i] - a[i - 1];
}
ll ans = gcd(d[2], d[3]);
for(int i = 4; i <= n; ++i)
{
ans = gcd(ans, d[i]);
}
for(int i = 1; i <= m; ++i)
{
scanf("%I64d", &p[i]);
}
ll y, pos;
bool flag = false;
for(int i = 1; i <= m; ++i)
{
if(ans % p[i] == 0)
{
pos = i;
flag = true;
break;
}
}
if(flag)
{
printf("YES\n");
printf("%I64d %I64d\n", a[1], pos);
}
else
printf("NO\n");
return 0;
}