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：给定一个斐波那契的前两位数求F（n）的后m位
思路：矩阵快速幂
AC代码：
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
using namespace std;
#define ll long long
ll mod;
const int N=2;
int tmp[N][N];
void multi(int a[][N],int b[][N],int n)
{
memset(tmp,0,sizeof tmp);
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
for(int k=0;k<n;k++)
{
tmp[i][j]=(tmp[i][j]+a[i][k]*b[k][j])%mod;
}
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
a[i][j]=tmp[i][j]%mod;
}
int res[N][N];
void Pow(int a[][N],int n)
{
memset(res,0,sizeof res);
for(int i=0;i<N;i++) res[i][i]=1;
while(n)
{
if(n&1)
multi(res,a,N);
multi(a,a,N);
n>>=1;
}
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int a,b,m;
ll n;
scanf("%d%d%lld%d",&a,&b,&n,&m);
mod=1;
while(m--)
{
mod*=10;
}
if(n==0)
{
printf("%d\n",a%mod);continue;
}
if(n==1)
{
printf("%d\n",b%mod);continue;
}
int x[N][N];
int y[N][N];
x[0][0]=1;
x[1][0]=1;
x[0][1]=1;
x[1][1]=0;
y[0][0]=b;
y[1][0]=a;
Pow(x,n-1);
multi(res,y,2);
printf("%d\n",res[0][0]);
}
}
这该死的pow函数，开方有误差可以理解，为什么次方还有误差，卡了我两个小时。。。。
注：pow大于一万的时候产生为1的误差