poj 2769 Reduced ID Numbers 同余定理
链接:http://poj.org/problem?id=2769
题意:寻找数m,是的对于n个数的余数不同
思路:暴力,优化:同余部分不用测试
代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <iostream> 2 #include <string.h> 3 using namespace std; 4 int ans[1000001],vis[1000001]; 5 int main() { 6 ios::sync_with_stdio(false); 7 //freopen("in.txt","r",stdin); 8 //freopen("out.txt","w",stdout); 9 int t,n,i,j,ok; 10 cin>>t; 11 while(t--) { 12 cin>>n; 13 for(i=1; i<=n; ++i) cin>>ans[i]; 14 for(i=1; i<1000001; ++i) { 15 ok=1; 16 for(j=0; j<=i-1; ++j) vis[j]=0; 17 for(j=1; j<=n; ++j) { 18 if(vis[ans[j]%i]) { 19 ok=0; 20 break; 21 } 22 vis[ans[j]%i]=1; 23 } 24 if(ok) break; 25 } 26 cout<<i<<endl; 27 } 28 return 0; 29 }