题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1576
思路：设定mod=9973，由于gcd(B,mod）=1，所以B对mod的逆元B'是可求得，然后(A/B)%mod=(A*B')%mod=((A%mod)*B')%mod=(n*B')%mod;所以本题的关键就是要求B对mod的逆元。
代码如下：
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef unsigned int ui; 4 typedef long long ll; 5 typedef unsigned long long ull; 6 #define pf printf 7 #define mem(a,b) memset(a,b,sizeof(a)) 8 #define prime1 1e9+7 9 #define prime2 1e9+9 10 #define pi 3.14159265 11 #define lson l,mid,rt<<1 12 #define rson mid+1,r,rt<<1|1 13 #define scand(x) scanf("%llf",&x) 14 #define f(i,a,b) for(int i=a;i<=b;i++) 15 #define scan(a) scanf("%d",&a) 16 #define mp(a,b) make_pair((a),(b)) 17 #define P pair<int,int> 18 #define dbg(args) cout<<#args<<":"<<args<<endl; 19 #define inf 0x7ffffff 20 inline int read(){ 21 int ans=0,w=1; 22 char ch=getchar(); 23 while(!isdigit(ch)){if(ch=='-')w=-1;ch=getchar();} 24 while(isdigit(ch))ans=(ans<<3)+(ans<<1)+ch-'0',ch=getchar(); 25 return ans*w; 26 } 27 int n,m,t; 28 const int maxn=1e5+10; 29 const int mod=9973; 30 ll exgcd(ll a,ll b,ll &x,ll &y) 31 { 32 if(b==0) 33 { 34 x=1,y=0; 35 return a; 36 } 37 ll ans=exgcd(b,a%b,x,y); 38 ll tmp = x; 39 x=y; 40 y=tmp-(a/b)*y; 41 return ans; 42 } 43 int main() 44 { 45 //freopen("input.txt","r",stdin); 46 //freopen("output.txt","w",stdout); 47 std::ios::sync_with_stdio(false); 48 t=read(); 49 while(t--) 50 { 51 ll n,B; 52 scanf("%lld%lld",&n,&B); 53 ll x,y; 54 exgcd(B,mod,x,y); 55 x=(x%mod+mod)%mod;//B的逆元 56 pf("%lld ",(x*n)%mod); 57 } 58 }