快速幂取模模板 && 51nod 1013 3的幂的和

 #include <iostream>
 #include <cstdio>
 #include <cmath>
 #include <vector>
 #include <cstring>
 #include <string>
 #include <algorithm>
 #include <string>
 #include <set>
 #include <functional>
 #include <numeric>
 #include <sstream>
 #include <stack>
 #include <map>
 #include <queue>
 #pragma comment(linker, "/STACK:102400000,102400000")
 #define CL(arr, val)    memset(arr, val, sizeof(arr))

 #define ll long long
 #define inf 0x7f7f7f7f
 #define lc l,m,rt<<1
 #define rc m + 1,r,rt<<1|1
 #define pi acos(-1.0)

 #define L(x)    (x) << 1
 #define R(x)    (x) << 1 | 1
 #define MID(l, r)   (l + r) >> 1
 #define Min(x, y)   (x) < (y) ? (x) : (y)
 #define Max(x, y)   (x) < (y) ? (y) : (x)
 #define lson l,m,rt<<1
 #define rson m+1,r,rt<<1|1
 #define E(x)        (1 << (x))
 #define iabs(x)     (x) < 0 ? -(x) : (x)
 #define OUT(x)  printf("%I64d\n", x)
 #define lowbit(x)   (x)&(-x)
 #define Read()  freopen("a.txt", "r", stdin)
 #define Write() freopen("b.txt", "w", stdout);
 #define maxn 100010
 #define mod 1000000007
 using namespace std;

 ll quick_mod(ll a,ll b,ll c)
 {
     ll ans=;
     while(b)
     {
         if(b&)
         {
             ans=(ans*a)%c;
             b--;
         }
         b/=;
         a=(a*a)%c;
     }
     return ans;
 }
 int main()
 {
    // freopen("a.txt","r",stdin);
     ll a,b,c;
     scanf("%lld%lld%lld",&a,&b,&c);
     printf("%lld\n",quick_mod(a,b,c));
     return ;
 }

http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1013

这个题需要用到两次二分：第一次是在求3^n次方,第二次是求3^0+3^1+...3^n.

第一次二分就是上面的快速幂,第二次二分：

加入n=5       3^1+3^2+3^3+3^4+3^5 = 3^1+3^2 + 3^2*(3^1+3^2) +3^5

这样就可以递归求解了,跟矩阵快速幂类似。

 #include <iostream>
 #include <cstdio>
 #include <cmath>
 #include <vector>
 #include <cstring>
 #include <string>
 #include <algorithm>
 #include <string>
 #include <set>
 #include <functional>
 #include <numeric>
 #include <sstream>
 #include <stack>
 #include <map>
 #include <queue>
 #pragma comment(linker, "/STACK:102400000,102400000")
 #define CL(arr, val)    memset(arr, val, sizeof(arr))

 #define ll long long
 #define inf 0x7f7f7f7f
 #define lc l,m,rt<<1
 #define rc m + 1,r,rt<<1|1
 #define pi acos(-1.0)

 #define L(x)    (x) << 1
 #define R(x)    (x) << 1 | 1
 #define MID(l, r)   (l + r) >> 1
 #define Min(x, y)   (x) < (y) ? (x) : (y)
 #define Max(x, y)   (x) < (y) ? (y) : (x)
 #define lson l,m,rt<<1
 #define rson m+1,r,rt<<1|1
 #define E(x)        (1 << (x))
 #define iabs(x)     (x) < 0 ? -(x) : (x)
 #define OUT(x)  printf("%I64d\n", x)
 #define lowbit(x)   (x)&(-x)
 #define Read()  freopen("a.txt", "r", stdin)
 #define Write() freopen("b.txt", "w", stdout);
 #define maxn 100010
 #define mod 1000000007
 using namespace std;

 ll power(ll a,ll b)
 {
     ll ans=;
     while(b)
     {
         if(b&)
         {
             ans=(ans*a)%mod;
             b--;
         }
         b/=;
         a=(a*a)%mod;
     }
     return ans;
 }
 ll c;
 ll sum(ll a,ll k)
 {
     if(k == ) return a;
     c=sum(a,k>>);
     ll ans=(c+c*power(a,(k>>)))%mod;  //每次 计算出两项
     if(k&) ans=(ans+power(a,k))%mod;  //是奇数的话要加上最后那一项
     return ans;
 }
 int main()
 {
    //freopen("a.txt","r",stdin);
     ll n;
     scanf("%lld",&n);
     //printf("%lld\n",(power(3,20)-1)/2%mod);
     printf("%lld\n",((sum(,n)%mod))+);
     return ;
 }