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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