1 solutions
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0
#include<bits/stdc++.h> using namespace std; typedef long long LL; const int P=200907; LL qmi(int a,int k) { LL res=1%P; while(k) //快速幂模板 { if(k&1) res=res*a%P; a=(LL)a*a%P; k=k>>1; } return res; } int main() { int t; cin>>t; while(t--) { LL a,b,c,k; cin>>a>>b>>c>>k; if(a+c==2*b) //等差数列 { cout<<(a+(b-a)*(k-1))%P<<endl; } else //等比数列 { cout<<a*qmi(b/a,k-1)%P<<endl; } } return 0; }
- 1
Information
- ID
- 187
- Time
- 1000ms
- Memory
- 256MiB
- Difficulty
- 5
- Tags
- # Submissions
- 13
- Accepted
- 5
- Uploaded By