Hello Kiki (Chinese remainder theorem template)

Topic links: http://acm.hdu.edu.cn/showproblem.php?pid=3579 

 

 1 #include <iostream>
 2 #include <algorithm>
 3 
 4 typedef long long LL;
 5 
 6 using namespace std;
 7 
 8 
 9 
10 
11 LL ex_gcd(LL a,LL b,LL &x,LL &y){
12     if (b == 0){
13         x = 1;
14         y = 0;
15         return a;
16     }
17     LL gcd = ex_gcd(b,a%b,y,x);
18     y -= (a/b)*x;
19     return gcd;
20 }
21 
22 LL inv(LL t,LL p){   // t 关于 p 的逆元
23     LL x,y,d;
24     d = ex_gcd(t,p,x,y);
25     if (d == 1){
26         return (x%p+p)%p;
27     }else
28         return -1;
29 }
30 
31 
32 // n个方程： x = a[i](mod m[i])   (0<=i<n)
33 
34 LL china(int n,LL *a,LL *m){
35     LL M = 1,ret = 0;
36     for (int i=0;i<n;i++){
37         M *= m[i];
38     }
39     for (int i=0;i<n;i++){
40         LL w = M/m[i];
41         ret = (ret + inv(w,m[i]) * w * a[i]) % M;
42     }
43     return (ret + M) % M;
44 }
45 
46 LL CRT(LL b[],LL n[],int num){
47     bool flag = false;
48     LL n1 = n[0],n2,b1 = b[0],b2,bb,d,t,k,x,y;
49     for (int i=1;i<num;i++){
50         n2 = n[i],b2 = b[i];
51         bb = b2 - b1;
52         d = ex_gcd(n1,n2,x,y);
53         if (bb%d){
54             flag = true;
55             break;
56         }
57         k = bb / d * x;
58         t = n2 / d;
59         if (t<0)
60             t = -t;
61         k = (k % t + t) % t;
62         b1 = b1 + n1 * k;
63         n1 = n1 / d * n2;
64     }
65     if (flag)
66         return -1;
67     if (b1 == 0){
68         b1 = n1;
69     }
70     return b1;
71 }
72 
73 
74 int main(){
75     int t,num;
76     LL b[55],n[55];
77     scanf("%d",&t);
78     int cas = 1;
79     while (t--){
80         scanf("%d",&num);
81         for (int i=0;i<num;i++){
82             scanf("%lld",&n[i]);
83         }
84         for (int i=0;i<num;i++){
85             scanf("%lld",&b[i]);
86         }
87         printf("Case %d: %lld\n",cas++,CRT(b,n,num));
88     }
89     return 0;
90 }