"Hdu1573-x problem" expands Euclidean-congruence equations

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

The number of X in a positive integer less than or equal to N is satisfied:

X MoD a0 = B0

X MoD A1 = B1

......

X mod ai = Bi

(0<ai<=10)

Input: First behavior a positive integerT, indicating that there areTGroup test Data. First behavior of each set of test data two positive integersN,m (0 < N <= 1000,000,000, 0 < M <=), indicatingXless than or equalN, the arrayaand thebin each hasMan element. In the next two lines, each row hasMa positive integer, respectively, for theaand thebthe element in the.

Output: Corresponds to each set of inputs, outputting a positive integer in a separate row that represents the number of X that satisfies the condition.

Sample Input

3

10 3

1 2 3

0 1 2

100 7

3 4 5 6 7 8 9

1 2 3 4 5 6 7

10000 10

1 2 3 4 5 6 7 8 9 10

0 1 2 3 4 5 6 7 8 9

Sample Output

1

0

3

Problem: The bare topic of the same residual equation group. In particular, it is important to note that a positive integer is required, that is, 0 of the time to special treatment.

#include <cstdio>#include<cstdlib>#include<cstring>#include<iostream>using namespaceStd;typedefLong LongLL;ConstLL m= -; LL A[m],b[m]; ll EXGCD (ll U,ll v,ll&x,ll &y) {    if(v==0) {x=1; y=0;returnu;} Else{LL tx,ty; LL D=EXGCD (v,u%v,tx,ty); X=Ty; Y=tx-(u/v) *Ty; returnD; }}intMain () {//freopen ("a.in", "R", stdin); //freopen ("A.out", "w", stdout);LL T; scanf ("%i64d",&T);  while(t--) {LL n,m; BOOLbk=1; scanf ("%i64d%i64d",&n,&m);  for(LL i=1; i<=m;i++) scanf ("%i64d",&A[i]);  for(LL i=1; i<=m;i++) scanf ("%i64d",&B[i]);        LL a,b,c,g,x,tx,ty,a1,b1; A1=a[1],b1=b[1];  for(LL i=2; i<=m;i++) {A=A1; B=a[i]; c=b[i]-B1; G=EXGCD (a,b,tx,ty); if(c%g) {bk=0; Break;} X= ((tx*c/g)% (b/g) + (b/g))% (b/g); B1=a1*x+B1; A1=a1/g*A[i]; }        if(BK) {/*after the indefinite equation is a1x+b1=p, the x=0 b1=p is obtained.            Because x can be equal to 0:ans= (N-B1)/a1+1, if p=0, then b1=x=0, does not satisfy positive integers, so subtract. */LL ans=0; if(N&GT;=B1) ans= (N-B1)/a1+1; if(Ans && b1==0) ans--; printf ("%i64d\n", ans); }        Elseprintf"0\n"); }    return 0;}

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"Hdu1573-x problem" expands Euclidean-congruence equations