HDU 5793 A Boring Question (找規律 : 快速冪+乘法逆元)
阿新 • • 發佈：2017-07-12
cnblogs and ott miss 逆元 找規律 -- for while
∑0≤k1,k2,?km≤n∏1?j<m(kj+1kj)%1000000007=?
We define that (kj+1kj)=kj+1!kj!(kj+1?kj)! . And (kj+1kj)=0 while kj+1<kj.
You have to get the answer for each n
For example,if n=1,m=3,
When k1=0,k2=0,k3=0,(k2k1)(k3k2)=1;
Whenk1=0,k2=1,k3=0,(k2k1)(k3k2)=0;
Whenk1=1,k2=0,k3=0,(k2k1)(k3k2)=0;
Whenk1=1,k2=1,k3=0,(k2k1)(k3k2)=0;
Whenk1=0,k2=0,k3=1,(k2k1)(k3k2)=1;
Whenk1=0,k2=1,k3=1,(k2k1)(k3k2)=1;
Whenk1=1,k2=0,k3=1,(k2k1)(k3k2)=0;
Whenk1=1,k2=1,k3=1,(k2k1)(k3k2)=1
So the answer is 4. Input The first line of the input contains the only integer T,(1≤T≤10000)
Then T lines follow,the i-th line contains two integers n,m,(0≤n≤109,2≤m≤109)
Output For each n and m,output the answer in a single line. Sample Input 2 1 2 2 3 Sample Output 3 13 Author UESTC Source 2016 Multi-University Training Contest 6 Recommend wange2014
A Boring Question
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 865 Accepted Submission(s): 534
∑0≤k1,k2,?km≤n∏1?j<m(kj+1kj)%1000000007=?
We define that (kj+1kj)=kj+1!kj!(kj+1?kj)! . And (kj+1kj)=0 while kj+1<kj.
You have to get the answer for each n
For example,if n=1,m=3,
When k1=0,k2=0,k3=0,(k2k1)(k3k2)=1;
Whenk1=0,k2=1,k3=0,(k2k1)(k3k2)=0;
Whenk1=1,k2=0,k3=0,(k2k1)(k3k2)=0;
Whenk1=1,k2=1,k3=0,(k2k1)(k3k2)=0;
Whenk1=0,k2=0,k3=1,(k2k1)(k3k2)=1;
Whenk1=0,k2=1,k3=1,(k2k1)(k3k2)=1;
Whenk1=1,k2=0,k3=1,(k2k1)(k3k2)=0;
Whenk1=1,k2=1,k3=1,(k2k1)(k3k2)=1
So the answer is 4. Input The first line of the input contains the only integer T,(1≤T≤10000)
Then T lines follow,the i-th line contains two integers n,m,(0≤n≤109,2≤m≤109)
Output For each n and m,output the answer in a single line. Sample Input 2 1 2 2 3 Sample Output 3 13 Author UESTC Source 2016 Multi-University Training Contest 6 Recommend wange2014
題解：
找規律...
f(0,2)=1; f(1,2)=3; f(2,2)=7; f(3,2)=15
f(0,3)=1; f(1,3)=4; f(2,3)=13;
f(0,4)=1; f(1,4)=5; f(2,4)=21;
f(0,5)=1; f(1,5)=6; f(2,5)=31;
f(n,m)=f(n-1,m)*m+1
所以 f(n,m)=(m^(n+1)-1)/(m-1)
//#include <iostream> #include<bits/stdc++.h> using namespace std; const long long mod=1000000007; long long ans,n,m; int T; long long poww(long long a,long long b) { long long ans=1; while(b) { if (b%2==1) ans=(ans*a)%mod; a=(a*a)%mod; b/=2; } return ans; } int main() { scanf("%d",&T); for(;T>0;T--) { scanf("%lld%lld",&n,&m); long long ans=poww(m,n+1); ans=(ans+mod-1)%mod; ans=(ans*poww(m-1,mod-2))%mod; printf("%lld\n",ans); } return 0; }
HDU 5793 A Boring Question (找規律 : 快速冪+乘法逆元)