标签:generator targe net name ima 技术 get lan strlen
题意:
思路:只要求出矩阵{{a,b}{1,0}}的n-1次方就能得出答案。学习了网上的十倍快速幂https://blog.csdn.net/To_the_beginning/article/details/88367974。
#include<cstdio> #include<cstring> #include<algorithm> #include<iostream> #include<map> #include<vector> #include<queue> #include<cmath> #define ll long long using namespace std; const int N = 1e6 + 10; struct node { ll mat[2][2]; }; ll x0, x1, a, b, mod; node ans, O, tmp, z; char n[N]; node cul (node x, node y) { for(int i = 0; i < 2; i++) for(int j = 0; j < 2; j++) { z.mat[i][j] = 0; for(int k = 0 ; k < 2; k++) z.mat[i][j] = (z.mat[i][j] + x.mat[i][k] * y.mat[k][j] ) % mod; } return z; } int main() { int op; scanf("%lld %lld %lld %lld %s %lld", &x0, &x1, &a, &b, n, &mod); int len = strlen(n); ans.mat[0][0] = ans.mat[1][1] = 1; tmp.mat[0][0] = a; tmp.mat[0][1] = b; tmp.mat[1][0] = 1; for(int i = len - 1; i >= 0; i--) { op = n[i] - ‘0‘; for(int j = 0; j < op; j++) ans = cul(ans, tmp); O.mat[0][0] = O.mat[1][1] = 1; O.mat[0][1] = O.mat[1][0] = 0; for(int j = 0; j < 10; j++) O = cul(O, tmp); tmp = O; } printf("%lld\n", (x1 * ans.mat[1][0] + x0 * ans.mat[1][1] ) % mod) ; return 0; }
标签:generator targe net name ima 技术 get lan strlen
原文地址:https://www.cnblogs.com/2462478392Lee/p/11290710.html