标签:快速幂 递推公式 scanf targe get line long target inline
题意:递推公式 Fn = Fn-1 + 2 * Fn-2 + n*n,让求 Fn;
析:很明显的矩阵快速幂,因为这个很像Fibonacci数列,所以我们考虑是矩阵,然后我们进行推公式,因为这样我们是无法进行运算的。好像有的思路,最后也没想出来,还是参考的大牛的博客
http://blog.csdn.net/spring371327/article/details/52973534
那是讲的很详细了,就不多说了,注意这个取模不是1e9+7,一开始忘了。。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #define debug puts("+++++") //#include <tr1/unordered_map> #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; //using namespace std :: tr1; typedef long long LL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const LL LNF = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1e5 + 5; const LL mod = 2147493647; const int N = 1e6 + 5; const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1}; const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1}; const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; inline LL gcd(LL a, LL b){ return b == 0 ? a : gcd(b, a%b); } inline int gcd(int a, int b){ return b == 0 ? a : gcd(b, a%b); } inline int lcm(int a, int b){ return a * b / gcd(a, b); } int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline int Min(int a, int b){ return a < b ? a : b; } inline int Max(int a, int b){ return a > b ? a : b; } inline LL Min(LL a, LL b){ return a < b ? a : b; } inline LL Max(LL a, LL b){ return a > b ? a : b; } inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } struct Matrix{ LL a[7][7]; Matrix operator * (const Matrix &p){ Matrix res; for(int i = 0; i < 7; ++i) for(int j = 0; j < 7; ++j){ res.a[i][j] = 0; for(int k = 0; k < 7; ++k) res.a[i][j] = (res.a[i][j] + a[i][k] * p.a[k][j]) % mod; } return res; } }; Matrix quick_pow(Matrix b, LL n){ Matrix res; memset(res.a, 0, sizeof res.a); for(int i = 0; i < 7; ++i) res.a[i][i] = 1; while(n){ if(n & 1) res = res * b; b = b * b; n >>= 1; } return res; } int main(){ Matrix x; memset(x.a, 0, sizeof x.a); x.a[0][0] = 1; x.a[0][1] = 2; x.a[0][2] = 1; x.a[0][3] = 4; x.a[0][4] = 6; x.a[0][5] = 4; x.a[0][6] = 1; x.a[1][0] = 1; x.a[2][2] = 1; x.a[2][3] = 4; x.a[2][4] = 6; x.a[2][5] = 4; x.a[2][6] = 1; x.a[3][3] = 1; x.a[3][4] = 3; x.a[3][5] = 3; x.a[3][6] = 1; x.a[4][4] = 1; x.a[4][5] = 2; x.a[4][6] = 1; x.a[5][5] = 1; x.a[5][6] = 1; x.a[6][6] = 1; int T; cin >> T; while(T--){ LL n, a, b; scanf("%I64d %I64d %I64d", &n, &a, &b); if(1 == n) printf("%I64d\n", a); else if(2 == n) printf("%I64d\n", b); else{ Matrix res = quick_pow(x, n-2); LL ans = 0; ans = (ans + res.a[0][0] * b) % mod; ans = (ans + res.a[0][1] * a) % mod; ans = (ans + res.a[0][2] * 16) % mod; ans = (ans + res.a[0][3] * 8) % mod; ans = (ans + res.a[0][4] * 4) % mod; ans = (ans + res.a[0][5] * 2) % mod; ans = (ans + res.a[0][6]) % mod; printf("%I64d\n", ans); } } return 0; }
5950 Recursive sequence (矩阵快速幂)
标签:快速幂 递推公式 scanf targe get line long target inline
原文地址:http://www.cnblogs.com/dwtfukgv/p/6024708.html