时间限制：1 秒

内存限制：32 兆

特殊判题：否

提交：6966

解决：978

//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 0x7fffffffusing 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;  }