HDU2276 Kiki & Little Kiki 2 矩阵应用

/* s[i][j]表示i秒钟后第j个灯的情况 s[i][j] = (s[i-1][j] + s[i-1][j-1]) % 2; 令 Fn(1,len) = |s[i][0] s[i][1] …… s[i][len]| F0(1,len)为输入状态 A(len len) = |1 1 | | 1 1 | | …… | |1 1| Fn = F0 * A ^ n 这题中模板会超时，要把模板中Multiply()的%放到()外面才可以 */ #include "Mat.h" #include <iostream> using namespace std; int main() { Mat A, F; mod = 2; char str[MAX]; int n, len, i; while (scanf("%d",&n)!=EOF) { scanf("%s",str); len = strlen(str); F.ReSize(1, len); A.ReSize(len, len); for(i = 0; i < len; i++) { F.s[0][i] = str[i]-'0'; A.s[i][i] = 1; if(i==0) A.s[len-1][i] = 1; else A.s[i-1][i] = 1; } A.Er_work(n); F.Multiply(A); for(i = 0; i < len; i++) printf("%d",F.s[0][i]); printf("\n"); } return 0; }