[51NOD1126]求递推序列的第n项(矩阵快速幂)

题目链接：http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1126

存在参数a,b为负数的情况。这时候要这么处理:

根据mod值（7）加至刚好大于0。

否则有些样例是过不去的。

 #include <algorithm>
 #include <iostream>
 #include <iomanip>
 #include <cstring>
 #include <climits>
 #include <complex>
 #include <fstream>
 #include <cassert>
 #include <cstdio>
 #include <bitset>
 #include <vector>
 #include <deque>
 #include <queue>
 #include <stack>
 #include <ctime>
 #include <set>
 #include <map>
 #include <cmath>

 using namespace std;

 const int maxn = ;
 typedef struct Matrix {
     int m[maxn][maxn];
     int r;
     int c;
     Matrix(){
         r = c = ;
         memset(m, , sizeof(m));
     }
 } Matrix;

 Matrix mul(Matrix m1, Matrix m2, int mod) {
     Matrix ans = Matrix();
     ans.r = m1.r;
     ans.c = m2.c;
     for(int i = ; i <= m1.r; i++) {
         for(int j = ; j <= m2.r; j++) {
                for(int k = ; k <= m2.c; k++) {
                 if(m2.m[j][k] == ) continue;
                 ans.m[i][k] = ((ans.m[i][k] + m1.m[i][j] * m2.m[j][k] % mod) % mod) % mod;
             }
         }
     }
     return ans;
 }

 Matrix quickmul(Matrix m, int n, int mod) {
     Matrix ans = Matrix();
     for(int i = ; i <= m.r; i++) {
         ans.m[i][i]  = ;
     }
     ans.r = m.r;
     ans.c = m.c;
     while(n) {
         if(n & ) {
             ans = mul(m, ans, mod);
         }
         m = mul(m, m, mod);
         n >>= ;
     }
     return ans;
 }

 int a, b, n, m;

 int main() {
     // freopen("in", "r", stdin);
        m = ;
     while(~scanf("%d %d %d", &a, &b, &n)) {
         Matrix p;
         while(a < ) a += m;
         while(b < ) b += m;
         p.r = , p.c = ;
         p.m[][] = ;
         p.m[][] = ;
         if(n <= ) {
             printf("%d\n", p.m[-n+][] % m);
             continue;
         }
         Matrix s;
         s.r = s.c = ;
         s.m[][] = a, s.m[][] = b;
         s.m[][] = , s.m[][] = ;
         s = quickmul(s, n-, m);
         int ans = ;
         for(int i = ; i <= p.r; i++) {
             ans = (ans + (p.m[i][] * s.m[][i]) % m) % m;
         }
         printf("%d\n", ans % m);
     }
     return ;
 }