Hdu_1573_X problem (segmentation: China Surplus, hdu_1573_x Segmentation

Hdu_1573_X problem (segmentation: China Surplus, hdu_1573_x Segmentation
Returns the number of X integers less than or equal to N. X mod a [0] = B [0], X mod a [1] = B [1], X mod a [2] = B [2],…, X mod a [I] = B [I],… (0 <a [I] <= 10 ).

The first line of Input data is a positive integer T, indicating that there are T groups of test data. The first behavior of each group of test data is two positive integers N, M (0 <N <= 1000,000,000, 0 <M <= 10), indicating that X is less than or equal to N, the arrays a and B have M elements. In the next two rows, each row has M positive integers, which are elements in a and B respectively. Output corresponds to each input group, and a positive integer is Output in an independent row, indicating the number of X that meets the conditions.

Sample Input

310 31 2 30 1 2100 73 4 5 6 7 8 91 2 3 4 5 6 710000 101 2 3 4 5 6 7 8 9 100 1 2 3 4 5 6 7 8 9

Sample Output

103

In my opinion, it is a deformation Question of China's residual theorem, or a deformation question of lcm.
Each number that matches the answer is between 0 and N, and the distance is lcm. If lcm + x is set to the solution that meets all a [I], lcm + x also meets the condition.
Certificate:
(Lcm + x) % a [I] = B [I]
Lcm % a [I] = 0
(Lcm + x) % a [I] = lcm % a [I] + (lcm + x) % a [I] = B [I]
Set t = N % lcm
In this question, we can determine whether there is a solution between 0-t and t-lcm + t.

#include <iostream>#include<cstdio>using namespace std;#define ll long longint a[15];int b[15];int gcd(int a,int b){    return b==0?a:gcd(b,a%b);}int main(){    int t;    scanf("%d",&t);    while(t--)    {        int n,m;        int lcm=1;        cin>>n>>m;        for(int i=0;i<m;i++)        {            cin>>a[i];            lcm=a[i]/gcd(lcm,a[i])*lcm;        }        for(int i=0;i<m;i++)        {            cin>>b[i];        }        int r=n%lcm;        int cnt1=0;        for(int i=1;i<=r&&!cnt1;i++)        {            for(int j=0;j<m;j++)            {                if(i%a[j]!=b[j])                    break;                if(j==m-1)                {                    cnt1++;                }            }        }        int cnt2=0;        for(int i=r+1;i<=r+lcm&&!cnt2;i++)        {            for(int j=0;j<m;j++)            {                if(i%a[j]!=b[j])                    break;                if(j==m-1)                    cnt2+=n/lcm;            }        }        cout<<cnt1+cnt2<<endl;    }}