UVA

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 andb, you can get many different sequences. Given the values of a, b, you have to find the last m digits off(n).

Input   The first line gives the number of test cases, which is less than 10001. Each test case consists of asingle 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

#include <iostream>#include <cstdio>#include <cstdlib>#include <string>#include <cmath>#include <algorithm>#include <cstring>#include <map>#include <sstream>#include <queue>#include <stack>#include <vector>using namespace std;#define INF 0x3f3f3f3f#define Mem(a,x) memset(a,x,sizeof(x))#define For(i,a,b) for(int i = a; i<b; i++)#define ll long long#define MAX_N 100010typedef vector<int> vec;typedef vector<vec> mat;int m;int mod[4] = {10,100,1000,10000};mat mul(mat &A,mat &B){    mat C(A.size(),vec(B[0].size()));    for(int i = 0; i<A.size(); i++)    {        for(int k = 0; k<B.size(); k++)        {            for(int j = 0; j<B[0].size(); j++)            {                C[i][j] = C[i][j] + A[i][k]*B[k][j];                C[i][j] %= mod[m-1];            }        }    }    return C;}mat pow(mat A,ll n){    mat B(A.size(),vec(A.size()));    for(int i = 0; i<A.size(); i++)        B[i][i] = 1;    while(n > 0)    {        if(n & 1) B = mul(B,A);        A = mul(A,A);        n >>= 1;    }    return B;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        int x[10];        Mem(x,0);        int a,b;        ll n;        scanf("%d%d%lld%d",&a,&b,&n,&m);        mat A(2,vec(2));        A[0][0] = 1,A[0][1] = 1;        A[1][0] = 1,A[1][1] = 0;        A = pow(A,n);        mat ans(2,vec(1));        ans[0][0] = b,ans[1][0] = a;        ans = mul(A,ans);        printf("%d\n",ans[1][0]);    }    return 0;}