hdu 2276 Kiki & Little Kiki 2

http://acm.hdu.edu.cn/showproblem.php?pid=2276

　　还是矩阵快速幂的题，将异或的操作转换成矩阵乘法。1y！

代码如下：

#include <cstdio>
#include <cstring>
#include <cassert>
#include <algorithm>

using namespace std;

const int maxSize = 100;
const int initMod = 1E9 + 7;
int curSize = maxSize;
int curMod = initMod;

struct Matrix {
    int val[maxSize][maxSize];

    Matrix(bool ONE = false) {
        for (int i = 0; i < curSize; i++) {
            for (int j = 0; j < curSize; j++) {
                val[i][j] = 0;
            }
            if (ONE) val[i][i] = 1;
        }
    }

    void print(int _l = curSize, int _w = curSize) {
        for (int i = 0; i < _l; i++) {
            for (int j = 0; j < _w; j++) {
                if (j) putchar(' ');
                printf("%d", val[i][j]);
            }
            puts("");
        }
        puts("~~");
    }
};

Matrix operator * (Matrix &_a, Matrix &_b) {
    Matrix ret = Matrix();

    for (int i = 0; i < curSize; i++) {
        for (int k = 0; k < curSize; k++) {
            if (_a.val[i][k]) {
                for (int j = 0; j < curSize; j++) {
                    ret.val[i][j] += _a.val[i][k] * _b.val[k][j];
                    ret.val[i][j] %= curMod;
                }
            }
        }
    }

    return ret;
}

Matrix operator ^ (Matrix &_a, int _p) {
    Matrix __a = _a, ret = Matrix(true);

    while (_p) {
        if (_p & 1) {
            ret = ret * __a;
        }
        __a = __a * __a;
        _p >>= 1;
    }

    return ret;
}

char *deal(int n) {
    char *buf = new char[101];
    int p = 0;

    scanf("%s", buf);
    curSize = strlen(buf);

    Matrix Base = Matrix(), op = Matrix(true);

    while (buf[p]) {
        Base.val[0][p] = buf[p] - '0';
        p++;
    }
    for (int i = 1; i < curSize; i++) {
        op.val[i - 1][i] = 1;
    }
    op.val[curSize - 1][0] = 1;
//    op.print();

    op = op ^ n;
    Base = Base * op;

    p = 0;
    while (buf[p]) {
        buf[p] = Base.val[0][p] + '0';
        p++;
    }

    return buf;
}

int main() {
    int n;

    curMod = 2;
    while (~scanf("%d", &n)) {
        printf("%s\n", deal(n));
    }

    return 0;
}

——written by Lyon