等比数列二分求和(logn复杂度)

看完这个之后,感觉数学简直太厉害了 转载自:http://blog.csdn.net/acdreamers/article/details/7851144

 

今天我们学习如何有效地求表达式的值。对于这个问题，用二分解决比较好。

 

（1）当时，

（2）当时，那么有

    

（3）当时，那么有

    

 

代码：

 

#include <iostream>  
#include <string.h>  
#include <stdio.h>  
  
using namespace std;  
const int M = 1000000007;  
typedef long long LL;  
  
LL power(LL a,LL b)  
{  
    LL ans = 1;  
    a %= M;  
    while(b)  
    {  
        if(b & 1)  
        {  
            ans = ans * a % M;  
            b--;  
        }  
        b >>= 1;  
        a = a * a % M;  
    }  
    return ans;  
}  
  
LL sum(LL a,LL n)  
{  
    if(n == 1) return a;  
    LL t = sum(a,n/2);  
    if(n & 1)  
    {  
        LL cur = power(a,n/2+1);  
        t = (t + t * cur % M) % M;  
        t = (t + cur) % M;  
    }  
    else  
    {  
        LL cur = power(a,n/2);  
        t = (t + t * cur % M) % M;  
    }  
    return t;  
}  
  
int main()  
{  
    LL a,n;  
    while(cin>>a>>n)  
        cout<<sum(a,n)<<endl;  
    return 0;  
}  

　　

题目：http://poj.org/problem?id=3233

 

题意：矩阵求和

 

代码：

 

#include <iostream>  
#include <string.h>  
#include <stdio.h>  
  
using namespace std;  
const int N = 35;  
  
struct Matrix  
{  
    int m[N][N];  
};  
  
Matrix I;  
int n,k,M;  
  
Matrix add(Matrix a,Matrix b)  
{  
    Matrix c;  
    for(int i=0; i<n; i++)  
    {  
        for(int j=0; j<n; j++)  
        {  
            c.m[i][j] = a.m[i][j] + b.m[i][j];  
            c.m[i][j] %= M;  
        }  
    }  
    return c;  
}  
  
Matrix multi(Matrix a,Matrix b)  
{  
    Matrix c;  
    for(int i=0; i<n; i++)  
    {  
        for(int j=0; j<n; j++)  
        {  
            c.m[i][j] = 0;  
            for(int k=0; k<n; k++)  
                c.m[i][j] += a.m[i][k] * b.m[k][j];  
            c.m[i][j] %= M;  
        }  
    }  
    return c;  
}  
  
Matrix power(Matrix A,int n)  
{  
    Matrix ans = I,p = A;  
    while(n)  
    {  
        if(n & 1)  
        {  
            ans = multi(ans,p);  
            n--;  
        }  
        n >>= 1;  
        p = multi(p,p);  
    }  
    return ans;  
}  
  
Matrix sum(Matrix A,int k)  
{  
    if(k == 1) return A;  
    Matrix t = sum(A,k/2);  
    if(k & 1)  
    {  
        Matrix cur = power(A,k/2+1);  
        t = add(t,multi(t,cur));  
        t = add(t,cur);  
    }  
    else  
    {  
        Matrix cur = power(A,k/2);  
        t = add(t,multi(t,cur));  
    }  
    return t;  
}  
  
int main()  
{  
    while(scanf("%d%d%d",&n,&k,&M)!=EOF)  
    {  
        Matrix A;  
        for(int i=0; i<n; i++)  
        {  
            for(int j=0; j<n; j++)  
            {  
                scanf("%d",&A.m[i][j]);  
                A.m[i][j] %= M;  
                I.m[i][j] = (i==j);  
            }  
        }  
        Matrix ans = sum(A,k);  
        for(int i=0; i<n; i++)  
        {  
            for(int j=0; j<n; j++)  
                printf("%d ",ans.m[i][j]);  
            puts("");  
        }  
    }  
    return 0;  
}