nylg 640 Geometric Sum

描述

Compute (a + a^2 + … + a^n) mod m.(a+a2+…an)mod

输入

Three integers a,n,m.
(1≤a,n,m≤10^18)
It ends with EOF.

输出

The only integer denotes the result.

样例输入

2 2 1000000000

样例输出

6 

来源

Lepus

 矩阵里也求过a+a^2+a^3+a^4.......

 1 #include<iostream>
 2 #include<stdio.h>
 3 #include<cstring>
 4 #include<cstdlib>
 5 #include<vector>
 6 using namespace std;
 7 typedef long long LL;
 8 
 9 LL sum_mod(LL a,LL n,LL p)
10 {
11     LL ans=0;
12     n=n%p;
13     while(n)
14     {
15         if(n&1)
16         {
17             ans=ans+a;
18             if(ans>=p) ans=ans-p;
19         }
20         n=n>>1;
21         a=(a+a)%p;
22     }
23     return ans;
24 }
25 LL solve(LL a,LL n,LL p)
26 {
27    LL p1=a,p2=a,ans,i;
28    vector<LL>Q;
29    while(n)
30    {
31        ans=(n&1);
32        Q.push_back(ans);
33        n=n>>1;
34    }
35    ans=Q.size()-2;
36    for(i=ans;i>=0;i--)
37    {
38        p1=sum_mod(p1,p2+1,p);
39        p2=sum_mod(p2,p2,p);
40        if(Q[i]==1)
41        {
42            p2=sum_mod(p2,a,p);
43            p1=(p1+p2)%p;
44        }
45    }
46    return p1;
47 }
48 int main()
49 {
50     LL a,n,p;
51     while(scanf("%lld%lld%lld",&a,&n,&p)>0)
52     {
53         if(n==1){
54             printf("%lld\n",a%p);
55             continue;
56         }
57         LL ans=solve(a,n,p);
58         printf("%lld\n",ans);
59     }
60     return 0;
61 }

 

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nylg 640 Geometric Sum

原文：http://www.cnblogs.com/tom987690183/p/3719687.html