POJ 2891 Strange Way to Express Integers

模线性同余方程组的求解

 

 1 #include <cstdio>

 2 #include <cstring>

 3 

 4 using namespace std;

 5 const int N = 1005;

 6 

 7 #define ll long long

 8 ll a[N] , b[N];

 9 

10 ll ex_gcd(ll a , ll &x , ll b , ll &y)

11 {

12     if(b == 0){

13         x = 1 , y = 0;

14         return a;

15     }

16     ll ans = ex_gcd(b , x , a%b , y);

17     ll t = x;

18     x= y , y = t - (a/b)*y;

19     return ans;

20 }

21 

22 ll mod_line(int n)

23 {

24     ll r = b[0] , lcm = a[0] , x , y;

25     for(int i = 1 ; i<n ; i++)

26     {

27         ll del = b[i] - r;

28         ll g = ex_gcd(lcm , x , a[i] , y);

29         if(del % g != 0) return -1;

30         ll Mod = a[i] / g;

31         x = ((x*del/g % Mod) + Mod)%Mod;

32         r = r + lcm*x;

33         lcm = lcm*a[i]/g;

34         r %= lcm;

35     }

36     return r;

37 }

38 

39 int main()

40 {

41    // freopen("a.in" , "r" , stdin);

42     int n;

43     while(scanf("%d" , &n) == 1)

44     {

45         for(int i= 0 ; i<n ; i++)

46             scanf("%I64d%I64d" , a+i , b+i);

47         printf("%I64d\n" , mod_line(n));

48     }

49     return 0;

50 }