X problem (Chinese remainder theorem + coprime version application) hdu1573

x Problem

Time limit:1000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 3921 Accepted Submission (s): 1253

Problem description The number of x in a positive integer less than or equal to n is satisfied: 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 behavior of input data is a positive integer t, which indicates that there is a T group of test data. The first behavior of each set 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, and that each has M elements in arrays A and B. The next two lines, each with m positive integers, are the elements in A and B, respectively.

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 Input310 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

In the application of Sample Output103, the Chinese remainder theorem (not coprime version) has just been learned; Test instructions will not speak, Chinese. Reprint please indicate source: http://www.cnblogs.com/yuyixingkong/topic Link: http://acm.hdu.edu.cn/showproblem.php?pid=1573

#include <stdio.h>#defineLL __int64LL a[ the],b[ the]; LL D,x,y,ans; LL DG;//Greatest Common DivisorvoidEXGCD (LL a,ll b,ll& d,ll& x,ll&y) {    if(!B) {d=a;x=1; y=0;} Else{EXGCD (b,a%b,d,y,x); Y-=x* (A/b); }}ll gcd (LL a,ll b) {if(!B)returnA; Elsegcd (B,a%b);}    LL China (ll N) {ll dm,a,b,d,x,y;    LL C1,c2,c; A=a[0]; C1=b[0];  for(intI=1; i<n;i++) {b=A[i]; C2=B[i];        EXGCD (A,b,d,x,y); DM=b/D; C=c2-C1; if(c%d)return-1; X= ((X*C/D)%DM+DM)%DM; C1=a*x+C1; A=a*b/D; } DG=A; if(c1==0) {C1=1;  for(intI=0; i<n;i++) {C1=c1*a[i]/gcd (C1,a[i]); } DG=C1; }    returnC1;}intMain () {LL t,n,m; scanf ("%i64d",&T);  while(t--) {scanf ("%i64d%i64d",&n,&M);  for(LL i=0; i<m;i++) scanf ("%i64d",&A[i]);  for(LL i=0; i<m;i++) scanf ("%i64d",&B[i]); Ans=China (M);//printf ("ans==%i64d\tdg=%i64d\n", ANS,DG);        if(ans==-1|| Ans>n) printf ("0\n"); Else{printf ("%i64d\n", (N-ans)/dg+1); }    }    return 0;}

X problem (Chinese remainder theorem + coprime version application) hdu1573