Hello Kiki
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1943 Accepted Submission(s): 693
Problem Description
One day I was shopping in the supermarket. There was a cashier counting coins seriously when a little kid running and singing "门前大桥下游过一群鸭，快来快来 数一数，二四六七八". And then the cashier put the counted coins back morosely and count again...
Hello Kiki is such a lovely girl that she loves doing counting in a different way. For example, when she is counting X coins, she count them N times. Each time she divide the coins into several same sized groups and write down the group size Mi and the number of the remaining coins Ai on her note.
One day Kiki's father found her note and he wanted to know how much coins Kiki was counting.
Hello Kiki is such a lovely girl that she loves doing counting in a different way. For example, when she is counting X coins, she count them N times. Each time she divide the coins into several same sized groups and write down the group size Mi and the number of the remaining coins Ai on her note.
One day Kiki's father found her note and he wanted to know how much coins Kiki was counting.
Input
The first line is T indicating the number of test cases.
Each case contains N on the first line, Mi(1 <= i <= N) on the second line, and corresponding Ai(1 <= i <= N) on the third line.
All numbers in the input and output are integers.
1 <= T <= 100, 1 <= N <= 6, 1 <= Mi <= 50, 0 <= Ai < Mi
Each case contains N on the first line, Mi(1 <= i <= N) on the second line, and corresponding Ai(1 <= i <= N) on the third line.
All numbers in the input and output are integers.
1 <= T <= 100, 1 <= N <= 6, 1 <= Mi <= 50, 0 <= Ai < Mi
Output
For each case output the least positive integer X which Kiki was counting in the sample output format. If there is no solution then output -1.
Sample Input
2
2
14 57
5 56
5
19 54 40 24 80
11 2 36 20 76
Sample Output
Case 1: 341
Case 2: 5996
Author
digiter (Special Thanks echo)
Source
Recommend
模板题，注意ans=0的情况，题目要求的是最小的正整数。
1 //0MS 244K 1571 B C++ 2 #include<stdio.h> 3 __int64 gcd(__int64 a,__int64 b) 4 { 5 return b?gcd(b,a%b):a; 6 } 7 __int64 extend_euclid(__int64 a,__int64 b,__int64 &x,__int64 &y) 8 { 9 if(b==0){ 10 x=1;y=0; 11 return a; 12 } 13 __int64 d=extend_euclid(b,a%b,x,y); 14 __int64 t=x; 15 x=y; 16 y=t-a/b*y; 17 return d; 18 } 19 __int64 inv(__int64 a,__int64 n) 20 { 21 __int64 x,y; 22 __int64 t=extend_euclid(a,n,x,y); 23 if(t!=1) return -1; 24 return (x%n+n)%n; 25 } 26 bool merge(__int64 a1,__int64 n1,__int64 a2,__int64 n2,__int64 &a3,__int64 &n3) 27 { 28 __int64 d=gcd(n1,n2); 29 __int64 c=a2-a1; 30 if(c%d) return false; 31 c=(c%n2+n2)%n2; 32 c/=d; 33 n1/=d; 34 n2/=d; 35 c*=inv(n1,n2); 36 c%=n2; 37 c*=n1*d; 38 c+=a1; 39 n3=n1*n2*d; 40 a3=(c%n3+n3)%n3; 41 return true; 42 } 43 __int64 china_reminder2(int len,__int64 *a,__int64 *n,__int64 &lcm) 44 { 45 __int64 a1=a[0],n1=n[0]; 46 __int64 a2,n2; 47 for(int i=1;i<len;i++){ 48 __int64 aa,nn; 49 a2=a[i],n2=n[i]; 50 if(!merge(a1,n1,a2,n2,aa,nn)) return -1; 51 a1=aa; 52 n1=nn; 53 } 54 lcm=n1; 55 return (a1%n1+n1)%n1; 56 } 57 int main(void) 58 { 59 int t,n,k=1; 60 __int64 a[10],b[10],lcm; 61 scanf("%d",&t); 62 while(t--) 63 { 64 scanf("%d",&n); 65 for(int i=0;i<n;i++) 66 scanf("%I64d",&a[i]); 67 for(int i=0;i<n;i++) 68 scanf("%I64d",&b[i]); 69 printf("Case %d: ",k++); 70 __int64 ans=china_reminder2(n,b,a,lcm); 71 if(ans==0) ans+=lcm; 72 printf("%I64d ",ans); 73 } 74 return 0; 75 }