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HDU 2276 Kiki & Little Kiki 2
矩阵快速幂。
0 1-> 第二个数字会变成1
0 0-> 第二个数字会变成0
1 0-> 第二个数字会变成1
1 1-> 第二个数字会变成0
根据这四个特点,就可以写转移矩阵了。
#pragma comment(linker, "/STACK:1024000000,1024000000")#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<iostream>using namespace std;typedef long long LL;const double pi=acos(-1.0),eps=1e-6;void File(){ freopen("D:\\in.txt","r",stdin); freopen("D:\\out.txt","w",stdout);}template <class T>inline void read(T &x){ char c = getchar(); x = 0; while(!isdigit(c)) c = getchar(); while(isdigit(c)) { x = x * 10 + c - ‘0‘; c = getchar(); }}char s[105];int len,m;struct Matrix{ int A[105][105]; int R, C; Matrix operator*(Matrix b);};Matrix X, Y, Z;Matrix ch(Matrix a, Matrix b){ Matrix c; memset(c.A, 0, sizeof(c.A)); int i, j, kk; for (i = 1; i <= len; i++) for (j = 1; j <= len; j++) { if(a.A[i][j]) { for (kk = 1; kk <= len; kk++) c.A[i][kk] = (c.A[i][kk]+a.A[i][j] * b.A[j][kk])%2; } } return c;}void init(){ for(int i=1;i<=len;i++) Z.A[1][i] = s[i-1]-‘0‘; Z.R = 1; Z.C = len; memset(Y.A,0,sizeof Y.A); for(int i=1;i<=len;i++) Y.A[i][i]=1; Y.R = len; Y.C = len; memset(X.A,0,sizeof X.A); X.A[1][1]=1; X.A[len][1]=1; for(int i=2;i<=len;i++) X.A[i][i]=1, X.A[i-1][i]=1; X.R = len; X.C = len;}void work(){ while (m) { if (m % 2 == 1) Y = ch(Y,X); m = m >> 1; X = ch(X,X); } Z = ch(Z,Y); for(int i=1;i<=len;i++) printf("%d",Z.A[1][i]); printf("\n");}int main(){ while(~scanf("%d%s",&m,s)) { len=strlen(s); init(); work(); } return 0;}
HDU 2276 Kiki & Little Kiki 2
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