题目链接：小数第n位
解题思路：因为求第n位开始的三位小数，那就要求到n+2，公式为： ( a / b ) ∗ 1 0 ( n + 2 ) % 1000 (a/b)*10^{(n+2)}\%1000 (a/b)∗10(n+2)%1000，例如1.2431的从4开始取为431等价于12431%1000=431
因为 x / d % m = x % ( d ∗ m ) / d x/d\%m=x\%(d*m)/d x/d%m=x%(d∗m)/d
所有 ( a / b ) ∗ 1 0 ( n + 2 ) % 1000 (a/b)*10^{(n+2)}\%1000 (a/b)∗10(n+2)%1000 => ( a ∗ 1 0 ( n + 2 ) ) % ( b ∗ 1000 ) / b (a *10 ^{ (n+2)})\%(b * 1000)/b (a∗10(n+2))%(b∗1000)/b
#include<bits/stdc++.h>
#define x first
#define y second
#define mem(h) memset(h,-1,sizeof h)
#define mcp(a,b) memcpy(a,b,sizeof b)
using namespace std;
typedef long long LL;
typedef unsigned long long ull;
typedef pair<int,int>PII;
typedef pair<double,double>PDD;
namespace IO{
inline LL read(){
LL o=0,f=1;char c=getchar();
while(c<'0'||c>'9'){
if(c=='-')f=-1;c=getchar();}
while(c>='0'&&c<='9'){
o=o*10+c-'0';c=getchar();}
return o*f;
}
}using namespace IO;
//#############以上是自定义技巧（可忽略）##########
const int N=1e2+7,M=2e5+7,INF=0x3f3f3f3f,mod=1e8+7,P=131;
LL a,b,n;
LL m;
LL Pow(LL Base,int M){
LL res=1;
while(M){
if(M&1)res=res*Base%m;
Base=Base*Base%m;
M>>=1;
}
return res;
}
int main(){
cin>>a>>b>>n;
m=1000*b;
LL res=Pow(10,n+2);
LL ans=(a*res)%m/b;
printf("%03d\n",ans);
return 0;
}