HDU 6050 17多校2 Funny Function(数学+乘法逆元)
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Problem Description
Function Fx,ysatisfies:
For given integers N and M,calculate Fm,1 modulo 1e9+7.
For given integers N and M,calculate Fm,1 modulo 1e9+7.
Input
There is one integer T in the first line.
The next T lines,each line includes two integers N and M .
1<=T<=10000,1<=N,M<2^63.
The next T lines,each line includes two integers N and M .
1<=T<=10000,1<=N,M<2^63.
Output
For each given N and M,print the answer in a single line.
Sample Input
2
2 2
3 3
Sample Output
2
33
数学题,反正我不会。。。数学基础太差了,直接用大佬的数学公式写的
参考博客:http://www.cnblogs.com/Just--Do--It/p/7248089.html
主要是学到了取余和除的话需要求逆元!!可以用费马小定理求,目前我只会这个
继续加油!
1 #include<cstdio> 2 #include<iostream> 3 #include<cstring> 4 #include<string.h> 5 #include<cmath> 6 #include<math.h> 7 using namespace std; 8 9 #define MOD 1000000000+7 10 11 long long quick_mod(long long a,long long b)//快速幂,复杂度log2n 12 { 13 long long ans=1; 14 while(b) 15 { 16 if(b&1) 17 { 18 ans*=a; 19 ans%=MOD; 20 b--; 21 } 22 b/=2; 23 a*=a; 24 a%=MOD; 25 } 26 return ans; 27 } 28 29 long long inv(long long x) 30 { 31 return quick_mod(x,MOD-2);//根据费马小定理求逆元,3*(3^(mod-2))=1 32 } 33 34 int main() 35 { 36 int T; 37 scanf("%d",&T); 38 long long n,m,ans; 39 while(T--) 40 { 41 scanf("%lld%lld",&n,&m); 42 if(n%2==0) 43 ans=(quick_mod(quick_mod(2,n)-1,m-1)*2)*inv(3); 44 else 45 ans=(quick_mod(quick_mod(2,n)-1,m-1)*2+1)*inv(3); 46 ans%=MOD; 47 printf("%lld\\n",ans); 48 } 49 return true; 50 }
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