Yet another Number Sequence

 Yet another Number Sequence

Let’s define another number sequence, given by the following function:
f(0) = a
f(1) = b
f(n) = f(n − 1) + f(n − 2), n > 1
When a = 0 and b = 1, this sequence gives the Fibonacci Sequence. Changing the values of a and
b, you can get many different sequences. Given the values of a, b, you have to find the last m digits of
f(n).
Input
The first line gives the number of test cases, which is less than 10001. Each test case consists of a
single line containing the integers a b n m. The values of a and b range in [0, 100], value of n ranges in
[0, 1000000000] and value of m ranges in [1, 4].
Output
For each test case, print the last m digits of f(n). However, you should NOT print any leading zero.
Sample Input
4
0 1 11 3
0 1 42 4
0 1 22 4
0 1 21 4
Sample Output
89
4296
7711
946

题意：给出a,b,n,m;a为第一个数，b为第二个数，然后以a，b为前两项a[n+1]=a[n]+a[n-1];构造数列

m为去几位数，即%10的几次方;n为此数列的第几个数

代码：

#include<iostream>
#include<cstdio>

using namespace std;

const int mod[5]={0,10,100,1000,10000};

#define ll long long

ll a,b,m,n;

struct matrix
{
    int map[2][2];
}ans,base;

matrix unit={1,0,0,1};

matrix multi(matrix a,matrix b)
{
    matrix c;
    for(int i=0;i<2;i++)
    {
        for(int j=0;j<2;j++)
        {
            c.map[i][j]=0;
            for(int k=0l;k<2;k++)
            {
                c.map[i][j]+=(a.map[i][k]*b.map[k][j])%mod[m];
            }
            c.map[i][j]%=mod[m];
        }
    }
    return c;
}

void pow(int n)
{
    base.map[0][0]=base.map[0][1]=base.map[1][0]=1;
    base.map[1][1]=0;
    ans.map[0][0]=ans.map[1][1]=1;
    ans.map[0][1]=ans.map[1][0]=0;
    while(n)
    {
        if(n&1)
            ans=multi(ans,base);
        base=multi(base,base);
        n>>=1;
    }
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lld%lld%lld%lld",&a,&b,&n,&m);
        if(n==0)
            return a;
        else if(n==1)
            return b;
        else
        {
            pow(n-1);
            ll result=(ans.map[0][0]*b+ans.map[0][1]*a)%mod[m];
            printf("%lld
",result);
        }
    }
    return 0;
}

//很有代表性