51 NOD 1013 3的幂的和

做法：快速幂+求逆元取模

因为ans=((3^(n+2))/2)%P

而ans%P/2!=ans/2%P

所以由费马小定理当gcd(a,p)==1&&P为质数时，a^(p-1)≡1（mod p）可得：ans*(p+1)/2≡ans/2 (%p)

然后就可以美滋滋地对ans取模辣

Code：

 1 #include <cstdio>
 2 inline int read()
 3 {
 4     register int f=1,k=0;register char c=getchar();
 5     while (c<'0'||c>'9')c=='-'&&(f=-1),c=getchar();
 6     while (c>='0'&&c<='9')k=k*10+c-'0',c=getchar();
 7     return k*f;
 8 }
 9 const long long MOD=1000000007;
10 int main()
11 {
12     register int n=read()+1;register long long t=3,ans=1;
13     while (n)
14     {
15         if(n&1)ans=ans*t%MOD;
16         t=t*t%MOD;
17         n>>=1;
18     }
19     printf("%lld
",ans); 
20     ans-=1;
21     printf("%lld
",ans*500000004%MOD);
22 }

View Code