说明：在本题中，答案是要求Fn除以10007的余数，因此我们只要能算出这个余数即可，而不需要先计算出Fn的准确值，再将计算的结果除以10007取余数，直接计算余数往往比先算出原数再取余简单。

 1 #include<cstdio> 2 #include<algorithm> 3 #include<cstring> 4 using namespace std; 5 const int maxn=1000000+10; 6 int f[maxn]; 7 int main() 8 { 9     int n;10     scanf("%d",&n);11     f[1]=1;12     f[2]=1;13     for(int i=2;i<=n;i++)14         f[i]=(f[i-1]+f[i-2])%10007;15     printf("%d\n",f[n]);16 17 }

 1 /* 2 ID: sdj22251 3 PROG: subset 4 LANG: C++ 5 */ 6 #include <iostream> 7 #include <vector> 8 #include <list> 9 #include <map>10 #include <set>11 #include <deque>12 #include <queue>13 #include <stack>14 #include <bitset>15 #include <algorithm>16 #include <functional>17 #include <numeric>18 #include <utility>19 #include <sstream>20 #include <iomanip>21 #include <cstdio>22 #include <cmath>23 #include <cstdlib>24 #include <cctype>25 #include <string>26 #include <cstring>27 #include <cmath>28 #include <ctime>29 #define MAXN 30530 #define INF 10000000031 #define eps 1e-732 #define PI 3.141592653589833 using namespace std;34 int n = 2, m;35 int tt[2][2]={{1, 1}, {1, 0}};36 struct wwj37 {38     int r, c;39     int mat[3][3];40 } need, ready;41 void init()42 {43     memset(need.mat, 0, sizeof(need.mat));44     need.r = n;45     need.c = n;46     for(int i = 1; i <= n; i++)47     {48         need.mat[i][i] = 1;49     }50     ready.c = n;51     ready.r = n;52     for(int i = 1; i <= n; i++)53     {54         for(int j = 1; j <= n; j++)55         ready.mat[i][j] = tt[i - 1][j - 1];56     }57 }58 wwj multi(wwj x, wwj y)59 {60     wwj t;61     int i, j, k;62     memset(t.mat, 0, sizeof(t.mat));63     t.r = x.r;64     t.c = y.c;65     for(i = 1; i <= x.r; i++)66     {67         for(k = 1; k <= x.c; k++)68             if(x.mat[i][k])69             {70                 for(j = 1; j <= y.c; j++)71                 {72                     t.mat[i][j] += (x.mat[i][k] * y.mat[k][j]) % 10007;73                     t.mat[i][j] %= 10007;74                 }75             }76     }77     return t;78 }79 int main()80 {81     int m;82     scanf("%d", &m);83         init();84         while(m)85         {86             if(m & 1)87             {88                 need = multi(ready, need);89             }90             ready = multi(ready, ready);91             m = m >> 1;92         }93         printf("%d\n", need.mat[1][2] % 10007);94     return 0;95 }