hdu 1005 Number Sequence

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原题链接：http://acm.hdu.edu.cn/showproblem.php?pid=1005
矩阵快速幂。。

#include<algorithm>#include<iostream>#include<cstdlib>#include<cstring>#include<cstdio>#include<vector>#include<map>using std::map;using std::min;using std::find;using std::pair;using std::vector;using std::multimap;#define pb(e) push_back(e)#define sz(c) (int)(c).size()#define mp(a, b) make_pair(a, b)#define all(c) (c).begin(), (c).end()#define iter(c) __typeof((c).begin())#define cls(arr, val) memset(arr, val, sizeof(arr))#define cpresent(c, e) (find(all(c), (e)) != (c).end())#define rep(i, n) for(int i = 0; i < (int)n; i++)#define tr(c, i) for(iter(c) i = (c).begin(); i != (c).end(); ++i)const int M = 7;const int N = 100001;const int INF = 0x3f3f3f3f;struct Matrix {    typedef vector<int> vec;    typedef vector<vec> mat;    inline mat mul(mat &A, mat &B) {        mat C(sz(A), vec(sz(B[0])));        rep(i, sz(A)) {            rep(k, sz(B)) {                rep(j, sz(B[0])) {                    C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % M;                }            }        }        return C;    }    inline mat pow(mat &A, int n) {        mat ret(sz(A), vec(sz(A[0])));        rep(i, sz(A)) ret[i][i] = 1;        while(n) {            if(n & 1) ret = mul(ret, A);            A = mul(A, A);            n >>= 1;        }        return ret;    }    inline void solve(int a, int b, int n) {        mat ans(2, vec(2));        ans[0][0] = a, ans[0][1] = b;        ans[1][0] = 1, ans[1][1] = 0;        ans = pow(ans, n - 2);        printf("%d\n", (ans[0][0] + ans[0][1]) % M);    }}go;int main() {#ifdef LOCAL    freopen("in.txt", "r", stdin);    freopen("out.txt", "w+", stdout);#endif    int a, b, n;    while(~scanf("%d %d %d", &a, &b, &n), a + b + n) {        if(1 == n || 2 == n) { puts("1"); continue; }        go.solve(a, b, n);    }    return 0;}

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