校赛学长出的一道题~
Last week, YaoBIG bought n lights from a shopping mall and he hoped to lighten all of them. Unfor-
tunately, there was something wrong with the lights. For the ith light, it just shone at the first ai minute
and re-shone every bi minutes. (This means: Except for the minutes ai, ai +bi, ai +2bi,..., the ith light
went out.)
YaoBIG just wanted to know if there was a time that all lights shone together?
Input
The input contains zero or more test cases and the first line of the input gives the number T of test
cases. For each test case:
The first line contains an integer n which represents the amount of the lights.
For the next n lines, there are two integers ai and bi in the ith line.
Output
For each test case, output the minimum time t when all lights shone together. If the time doesn’t
exist, just output -11.
Limits
0 <=T <=1 000
1 <=n<=300
0 <=ai <=bi<=16
1 #include<cstdio> 2 #include<cmath> 3 #include<bits/stdc++.h> 4 #define rep(i,a,b) for(int i=a;i<=b;++i) 5 typedef long long ll; 6 const int MAXN=310; 7 int a[MAXN];int b[MAXN]; 8 using namespace std; 9 ll extend_gcd(ll a,ll b,ll &x,ll &y) //扩展欧几里得算法，求gcd的同时递归修改x,y，最终满足xa+yb=gcd(a,b) 10 { 11 if(b==0) 12 { 13 x=1;y=0;return a; 14 } 15 ll r=extend_gcd(b,a%b,x,y); 16 ll tmp=x; 17 x=y;y=tmp-(a/b)*y; 18 return r; 19 } 20 class node 21 { 22 public: 23 ll a,b; 24 node(ll x=0,ll y=0):a(x),b(y){} 25 node operator +(const node &A)const 26 { 27 if(b==0||A.b==0) return node(0,0); //代表前面的方程组已经无解 28 ll x,y; 29 ll gcd=extend_gcd(b,A.b,x,y); 30 if((a-A.a)%gcd!=0) return node(0,0); //条件不满足则无解 31 ll newb=b*A.b/gcd,newa=newb; //核心部分，具体请看前面写的证明 32 x=(x*(A.a-a)*b/gcd)%newb; 33 x=(x+newb)%newb; 34 newa=(a+x)%newb; 35 return node(newa,newb); 36 } 37 38 }; 39 int main() 40 { 41 // freopen("in.txt","r",stdin); 42 int T; 43 scanf("%d",&T); 44 rep(k,1,T) 45 { 46 int n;scanf("%d",&n); 47 rep(i,1,n) scanf("%d%d",&a[i],&b[i]); 48 node tmp=node(0,1); 49 rep(i,1,n) tmp=tmp+node(a[i],b[i]); 50 printf("%lld ",tmp.b==0?-1:tmp.a); 51 } 52 return 0; 53 }