思路：很普通的矩阵快速幂，直接构造就可以了。但是这题贼坑，他所谓的取膜和我们平时说的不一样，负数取膜要变成正的，浪费了我2个多小时一直以为是模板问题。总之垃圾题一道。
1 #include <iostream> 2 #include <queue> 3 #include <stack> 4 #include <cstdio> 5 #include <vector> 6 #include <map> 7 #include <set> 8 #include <bitset> 9 #include <algorithm> 10 #include <cmath> 11 #include <cstring> 12 #include <cstdlib> 13 #include <string> 14 #include <sstream> 15 #include <time.h> 16 #define x first 17 #define y second 18 #define pb push_back 19 #define mp make_pair 20 #define lson l,m,rt*2 21 #define rson m+1,r,rt*2+1 22 #define mt(A,B) memset(A,B,sizeof(A)) 23 #define mod 7 24 using namespace std; 25 typedef long long LL; 26 const double PI = acos(-1); 27 const int N=2; 28 const int inf = 0x3f3f3f3f; 29 const LL INF=0x3f3f3f3f3f3f3f3fLL; 30 LL one,two,three; 31 struct Mat 32 { 33 LL mat[N][N]; 34 }; 35 Mat operator *(Mat a,Mat b) 36 { 37 Mat c; 38 memset(c.mat,0,sizeof(c.mat)); 39 int i,j,k; 40 for(i=0; i<N; i++) 41 { 42 for(j=0; j<N; j++) 43 { 44 for(k=0; k<N; k++) 45 { 46 c.mat[i][j]=(c.mat[i][j]+a.mat[i][k]*b.mat[k][j])%7; 47 } 48 } 49 } 50 return c; 51 } 52 Mat powermod(Mat a,int n) 53 { 54 Mat ans={1,0,1,0};//初始矩阵 55 int i,j; 56 if(n<0)return ans; 57 while(n) 58 { 59 if(n&1) 60 { 61 ans=ans*a;//矩阵a是我们所要构造的矩阵即a^n。 62 } 63 a=a*a; 64 n/=2; 65 } 66 return ans; 67 } 68 int main() 69 { 70 #ifdef Local 71 freopen("data.txt","r",stdin); 72 #endif 73 int n; 74 LL o; 75 cin>>one>>two>>n; 76 Mat p={0,(two%mod+mod)%mod,1,(one%mod+mod)%mod},q; 77 q=powermod(p,n-2); 78 cout<<q.mat[0][1]<<endl; 79 #ifdef Local 80 cerr << "time: " << (LL) clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl; 81 #endif 82 }