【题目大意】
给出k个模方程组：x mod ai = ri。求x的最小正值。如果不存在这样的x，那么输出-1.
【题解】
模板题，练习剩余定理的模板
1 /************* 2 POJ 2891 3 by chty 4 2016.11.3 5 *************/ 6 #include<iostream> 7 #include<cstdlib> 8 #include<cstdio> 9 #include<cstring> 10 #include<ctime> 11 #include<cmath> 12 #include<algorithm> 13 using namespace std; 14 #define MAXN 100000 15 typedef long long ll; 16 ll n,a[MAXN],m[MAXN]; 17 inline ll read() 18 { 19 ll x=0,f=1; char ch=getchar(); 20 while(!isdigit(ch)) {if(ch=='-') f=-1; ch=getchar();} 21 while(isdigit(ch)) {x=x*10+ch-'0'; ch=getchar();} 22 return x*f; 23 } 24 ll exgcd(ll a,ll b,ll &x,ll &y) 25 { 26 if(!b) {x=1; y=0; return a;} 27 ll r=exgcd(b,a%b,x,y); 28 ll t=x;x=y;y=t-a/b*y; 29 return r; 30 } 31 ll China() 32 { 33 ll A=a[1],M=m[1],k,y; 34 for(ll i=2;i<=n;i++) 35 { 36 ll g=exgcd(M,m[i],k,y),da=a[i]-A; 37 if(da%g) return -1; 38 ll t=m[i]/g; 39 k*=da/g; 40 k=(k%t+t)%t; 41 A+=k*M; 42 M=M*m[i]/g; 43 A=(A+M)%M; 44 } 45 return A; 46 } 47 int main() 48 { 49 freopen("cin.in","r",stdin); 50 freopen("cout.out","w",stdout); 51 while(~scanf("%lld",&n)) 52 { 53 for(ll i=1;i<=n;i++) m[i]=read(),a[i]=read(); 54 printf("%lld ",China()); 55 } 56 return 0; 57 }