标签：des style class blog code http
| Time Limit: 1000MS | Memory Limit: 131072K | |
| Total Submissions: 9472 | Accepted: 2873 |
Description
Elina is reading a book written by Rujia Liu, which introduces a strange way to express non-negative integers. The way is described as following:
Choose k different positive integers a1, a2, …, ak. For some non-negative m, divide it by every ai (1 ≤ i ≤ k) to find the remainder ri. If a1, a2, …, ak are properly chosen, m can be determined, then the pairs (ai, ri) can be used to express m.
“It is easy to calculate the pairs from m, ” said Elina. “But how can I find m from the pairs?”
Since Elina is new to programming, this problem is too difficult for her. Can you help her?
Input
The input contains multiple test cases. Each test cases consists of some lines.
Output
Output the non-negative integer m on a separate line for each test case. If there are multiple possible values, output the smallest one. If there are no possible values, output -1.
Sample Input
2 8 7 11 9
Sample Output
31
Hint
All integers in the input and the output are non-negative and can be represented by 64-bit integral types.
Source
1 //156K 16MS C++ 1362B 2014-06-13 12:36:23 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) 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 return (a1%n1+n1)%n1; 55 } 56 int main(void) 57 { 58 int n; 59 __int64 a[1005],b[1005]; 60 while(scanf("%d",&n)!=EOF) 61 { 62 for(int i=0;i<n;i++) 63 scanf("%I64d %I64d",&a[i],&b[i]); 64 printf("%I64d\n",china_reminder2(n,b,a)); 65 } 66 return 0; 67 }
poj 2891 Strange Way to Express Integers (非互质的中国剩余定理),布布扣,bubuko.com
poj 2891 Strange Way to Express Integers (非互质的中国剩余定理)
标签：des style class blog code http
原文地址：http://www.cnblogs.com/GO-NO-1/p/3785591.html
