每日一九度之 题目1081：递推数列

//Asimple
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cctype>
#include <cstdlib>
#include <stack>
#include <cmath>
#include <set>
#include <map>
#include <string>
#include <queue>
#include <limits.h>
#define INF 0x7fffffff
using namespace std;
const int maxn = 1005;
const int mod = 10000;
typedef long long ll;
int a[maxn];
int a0, a1, p, q, k;
int main(){
    while( ~scanf("%d %d %d %d %d",&a0, &a1, &p, &q, &k) ){
        a[0] = a0%mod;
        a[1] = a1%mod;
        for(int i=2; i<=k; i++){
            a[i] = (p*a[i-1]%mod + q*a[i-2]%mod)%mod;
        }
        printf("%d\n",a[k]);
    } 
    return 0;
}

#include <stdio.h>  
#include <stdlib.h>  
#define MOD 10000       //结果取MOD，避免高精度运算  
   
/*将矩阵p与矩阵q相乘，结果存入p矩阵*/  
void Matrix_mul(int p[2][2], int q[2][2])  
{  
    int i, j, k;  
    int t[2][2]={0};  
    for(i = 0; i <= 1; i++)  
        for(j = 0; j <= 1; j++)  
            for(k = 0; k <= 1; k++)  
                t[i][j] += p[i][k] * q[k][j];  
    for(i = 0; i <= 1; i++)  
        for(j = 0; j <= 1; j++)  
            p[i][j] = t[i][j] % MOD;  
}  
   
/*计算p矩阵的n次方，结果存入p矩阵*/  
void Matrix_cal(int p[2][2], int n)  
{  
    int i, j;  
    int t[2][2];  
    for(i = 0; i <= 1; i++)  
        for(j = 0; j <= 1; j++)  
            t[i][j] = p[i][j];  
    if(n == 1)   
        return;  
    else if(n & 1)  
    {  
        Matrix_cal(p, n-1);  
        Matrix_mul(p, t);  
    }  
    else  
    {  
        Matrix_cal(p, n/2);  
        Matrix_mul(p, p);  
    }  
}  
    
int main()  
{  
    int a0, a1, p, q, k;  
    while(scanf("%d%d%d%d%d", &a0, &a1, &p, &q, &k) != EOF)  
    {  
        if(k == 0)   
            printf("%d\n", a0);    
        else if(k == 1)   
            printf("%d\n", a1);   
        else  
        {  
            int matrix[2][2] = { {p%MOD, q%MOD}, {1, 0} };  
            Matrix_cal(matrix, k-1);  
            printf("%d\n", (a1 * matrix[0][0] + a0 * matrix[0][1]) % MOD);  
        }  
    }   
    return 0;  
}