Hdu1573x (Chinese Remainder Theorem of non-reciprocal elements)

Source: Internet
Author: User
X problems Time Limit: 1000/1000 MS (Java/others) memory limit: 32768/32768 K (Java/Others)
Total submission (s): 3295 accepted submission (s): 1068


Problem description calculates the number of X values in a positive integer 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
The number of M groups. Each group represents X % AI = Bi. Returns the number of X in the N range. Because all AI is not of mutual quality, we cannot directly use the Chinese residue theorem, but we can use the equation (for a * x + B * Y = C, X increases by B/GAD (a, B) each time, and then you only need to determine the number of occurrences within n.
#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define LL __int64LL t , n , m , d , x , y , i , bb , aa , flag ;void gcd(LL a,LL b){    if(b == 0)    {        d = a ; x = 1 ; y = 0 ;    }    else    {        gcd(b,a%b);        swap(x,y);        x = -x ; y = -y ;        y += (a/b)*x ;    }    return ;}LL a[30] , b[30] ;int main(){    scanf("%I64d", &t);    while(t--)    {        scanf("%I64d %I64d", &n, &m);        for(i = 0 ; i < m ; i++)            scanf("%I64d", &a[i]);       for(i = 0 ; i < m ; i++)            scanf("%I64d", &b[i]);        aa = a[0] ;        bb = b[0] ;        flag = 1 ;        for(i = 1 ; i < m ; i++)        {            gcd(aa,a[i]);            if( (b[i]-bb)%d != 0 )                flag = 0 ;            if( flag )            {                x = (b[i]-bb)/d*x ;                y = a[i] / d ;                x = ( x%y + y )%y ;                bb = bb + x * aa ;                aa = aa*a[i]/d ;            }        }        gcd(1,aa);        if( bb%d != 0 )            flag = 0 ;        if( flag )        {            x = ( bb/d )*x ;            y = aa / d ;            x = (x % y + y) % y ;        }        if( flag == 0 || x > n )            printf("0\n");        else        {            if( !x )                printf("%I64d\n", (n-x)/y );            else            printf("%I64d\n", (n-x)/y+1 );        }    }    return 0;}

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