uva 10862 uva 10334 uva 10359 uva 10918 （大数+递推）

10334是fib，不说了。

10862：

题意：

给n个房子，让他们和一个信号站连，求连法。

解析：

设f[n]代表n个房子的连法。

             O

房子堆n-1  房子n

到达当前决策n时，房子n可以连圈，则此时连法是f[n -1]，可以连房子堆n-1，连法也是f[n - 1]。

还有一种连法就是房子n与房子n-1，圈圈O都相连，此时影响到前面的决策，即房子n-2不能与O相连，则连法f[n -1 ] - f[n - 2]。

和：3 * f[n-1] - f[n-2]

代码：

#include <cstdio>#include <iostream>#include <cstring>#include <climits>#include <cassert>using namespace std;#define maxn 30000struct bign{    long long len, s[maxn];    bign()    {        memset(s, 0, sizeof(s));        len = 1;    }    bign(long long num)    {        *this = num;    }    bign(const char* num)    {        *this = num;    }    bign operator = (long long num)    {        char s[maxn];        sprintf(s, "%d", num);        *this = s;        return *this;    }    bign operator = (const char* num)    {        len = strlen(num);        for (long long i = 0; i < len; i++) s[i] = num[len-i-1] - '0';        return *this;    }    string str() const    {        string res = "";        for (long long i = 0; i < len; i++) res = (char)(s[i] + '0') + res;        if (res == "") res = "0";        return res;    }    /*去除前导0*/    void clean()    {        while(len > 1 && !s[len-1]) len--;    }    /*高精度的加法*/    bign operator + (const bign& b) const    {        bign c;        c.len = 0;        for (long long i = 0, g = 0; g || i < max(len, b.len); i++)        {            long long x = g;            if (i < len) x += s[i];            if (i < b.len) x += b.s[i];            c.s[c.len++] = x % 10;            g = x / 10;        }        return c;    }    /*高精度的减法*/    bign operator - (const bign& b)    {        bign c;        c.len = 0;        for (long long i = 0, g = 0; i < len; i++)        {            long long x = s[i] - g;            if (i < b.len) x -= b.s[i];            if (x >= 0)                g = 0;            else            {                g = 1;                x += 10;            }            c.s[c.len++] = x;        }        c.clean();        return c;    }    /*高精度的乘法*/    bign operator * (const bign& b)    {        bign c;        c.len = len + b.len;        for (long long i = 0; i < len; i++)            for (long long j = 0; j < b.len; j++)                c.s[i+j] += s[i] * b.s[j];        for (long long i = 0; i < c.len-1; i++)        {            c.s[i+1] += c.s[i] / 10;            c.s[i] %= 10;        }        c.clean();        return c;    }    /*高精度除以低精度*/    /*用到除法和取余的时候可能需要把全局的int替换成long long*/    bign operator / (long long b) const    {        assert(b > 0);        bign c;        c.len = len;        for (long long i = len-1, g = 0; i >= 0; --i)        {            long long x = 10*g+s[i];    //这里可能会超过int 故用long long            c.s[i] = x/b;                //这里可能会超过int            g = x-c.s[i]*b;                //这里可能会超过int        }        c.clean();        return c;    }    /*高精度对低精度取余*/    /*用到除法和取余的时候可能需要把全局的int替换成long long*/    bign operator % (long long b)    {        assert(b > 0);        bign d = b;        bign c = *this-*this/b*d;        return c;    }    bool operator < (const bign& b) const    {        if (len != b.len) return len < b.len;        for (long long i = len-1; i >= 0; i--)            if (s[i] != b.s[i]) return s[i] < b.s[i];        return false;    }    bool operator > (const bign& b) const    {        return b < *this;    }    bool operator <= (const bign& b)    {        return !(b > *this);    }    bool operator >= (const bign& b)    {        return !(b < *this);    }    bool operator == (const bign& b)    {        return !(b < *this) && !(*this < b);    }    bool operator != (const bign& b)    {        return (b < *this) || (*this < b);    }    bign operator += (const bign& b)    {        *this = *this + b;        return *this;    }};istream& operator >> (istream &in, bign& x){    string s;    in >> s;    x = s.c_str();    return in;}ostream& operator << (ostream &out, const bign& x){    out << x.str();    return out;}bign f[2000 + 10];void init(){    f[1] = 1, f[2] = 3;    bign three = 3;    for (int i = 3; i <= 2000; i++)        f[i] = three * f[i - 1] - f[i - 2];}int main(){#ifdef LOCAL    freopen("in.txt", "r" , stdin);#endif    init();    int n;    while (cin >> n && n)        cout << f[n] << endl;}

10359：

题意：

用1*1和1*2的瓷砖，问能贴几个n*2的方格。

解析：

当前格，加1格得到：f（n-1）种方法，加2格得到：2 * f（n-2）种方法。

0的时候输出1，。。。。。。。。。。。。。。。。。

代码：

#include <cstdio>#include <iostream>#include <cstring>#include <climits>#include <cassert>using namespace std;#define maxn 30000struct bign{    long long len, s[maxn];    bign()    {        memset(s, 0, sizeof(s));        len = 1;    }    bign(long long num)    {        *this = num;    }    bign(const char* num)    {        *this = num;    }    bign operator = (long long num)    {        char s[maxn];        sprintf(s, "%d", num);        *this = s;        return *this;    }    bign operator = (const char* num)    {        len = strlen(num);        for (long long i = 0; i < len; i++) s[i] = num[len-i-1] - '0';        return *this;    }    string str() const    {        string res = "";        for (long long i = 0; i < len; i++) res = (char)(s[i] + '0') + res;        if (res == "") res = "0";        return res;    }    /*去除前导0*/    void clean()    {        while(len > 1 && !s[len-1]) len--;    }    /*高精度的加法*/    bign operator + (const bign& b) const    {        bign c;        c.len = 0;        for (long long i = 0, g = 0; g || i < max(len, b.len); i++)        {            long long x = g;            if (i < len) x += s[i];            if (i < b.len) x += b.s[i];            c.s[c.len++] = x % 10;            g = x / 10;        }        return c;    }    /*高精度的减法*/    bign operator - (const bign& b)    {        bign c;        c.len = 0;        for (long long i = 0, g = 0; i < len; i++)        {            long long x = s[i] - g;            if (i < b.len) x -= b.s[i];            if (x >= 0)                g = 0;            else            {                g = 1;                x += 10;            }            c.s[c.len++] = x;        }        c.clean();        return c;    }    /*高精度的乘法*/    bign operator * (const bign& b)    {        bign c;        c.len = len + b.len;        for (long long i = 0; i < len; i++)            for (long long j = 0; j < b.len; j++)                c.s[i+j] += s[i] * b.s[j];        for (long long i = 0; i < c.len-1; i++)        {            c.s[i+1] += c.s[i] / 10;            c.s[i] %= 10;        }        c.clean();        return c;    }    /*高精度除以低精度*/    /*用到除法和取余的时候可能需要把全局的int替换成long long*/    bign operator / (long long b) const    {        assert(b > 0);        bign c;        c.len = len;        for (long long i = len-1, g = 0; i >= 0; --i)        {            long long x = 10*g+s[i];    //这里可能会超过int 故用long long            c.s[i] = x/b;                //这里可能会超过int            g = x-c.s[i]*b;                //这里可能会超过int        }        c.clean();        return c;    }    /*高精度对低精度取余*/    /*用到除法和取余的时候可能需要把全局的int替换成long long*/    bign operator % (long long b)    {        assert(b > 0);        bign d = b;        bign c = *this-*this/b*d;        return c;    }    bool operator < (const bign& b) const    {        if (len != b.len) return len < b.len;        for (long long i = len-1; i >= 0; i--)            if (s[i] != b.s[i]) return s[i] < b.s[i];        return false;    }    bool operator > (const bign& b) const    {        return b < *this;    }    bool operator <= (const bign& b)    {        return !(b > *this);    }    bool operator >= (const bign& b)    {        return !(b < *this);    }    bool operator == (const bign& b)    {        return !(b < *this) && !(*this < b);    }    bool operator != (const bign& b)    {        return (b < *this) || (*this < b);    }    bign operator += (const bign& b)    {        *this = *this + b;        return *this;    }};istream& operator >> (istream &in, bign& x){    string s;    in >> s;    x = s.c_str();    return in;}ostream& operator << (ostream &out, const bign& x){    out << x.str();    return out;}bign f[250 + 10];void init(){    f[0] = 1;    f[1] = 1, f[2] = 3;    bign two = 2;    for (int i = 3; i <= 250; i++)        f[i] = two * f[i - 2] + f[i - 1];}int main(){#ifdef LOCAL    freopen("in.txt", "r" , stdin);#endif    init();    int n;    while (cin >> n)        cout << f[n] << endl;}

10918：

题意：

给n，求用1*2的瓷砖有多少种方法能谱n*3的矩形。

解析：

奇数无解，当前选择，由n-2的瓷砖得来，有3 * f（n-2）种方法，若由n-4的瓷砖得来（f（n-2） - f（n-4））种方法。

代码：

#include <cstdio>#include <iostream>#include <cstring>#include <climits>#include <cassert>using namespace std;#define maxn 30000struct bign{    long long len, s[maxn];    bign()    {        memset(s, 0, sizeof(s));        len = 1;    }    bign(long long num)    {        *this = num;    }    bign(const char* num)    {        *this = num;    }    bign operator = (long long num)    {        char s[maxn];        sprintf(s, "%d", num);        *this = s;        return *this;    }    bign operator = (const char* num)    {        len = strlen(num);        for (long long i = 0; i < len; i++) s[i] = num[len-i-1] - '0';        return *this;    }    string str() const    {        string res = "";        for (long long i = 0; i < len; i++) res = (char)(s[i] + '0') + res;        if (res == "") res = "0";        return res;    }    /*去除前导0*/    void clean()    {        while(len > 1 && !s[len-1]) len--;    }    /*高精度的加法*/    bign operator + (const bign& b) const    {        bign c;        c.len = 0;        for (long long i = 0, g = 0; g || i < max(len, b.len); i++)        {            long long x = g;            if (i < len) x += s[i];            if (i < b.len) x += b.s[i];            c.s[c.len++] = x % 10;            g = x / 10;        }        return c;    }    /*高精度的减法*/    bign operator - (const bign& b)    {        bign c;        c.len = 0;        for (long long i = 0, g = 0; i < len; i++)        {            long long x = s[i] - g;            if (i < b.len) x -= b.s[i];            if (x >= 0)                g = 0;            else            {                g = 1;                x += 10;            }            c.s[c.len++] = x;        }        c.clean();        return c;    }    /*高精度的乘法*/    bign operator * (const bign& b)    {        bign c;        c.len = len + b.len;        for (long long i = 0; i < len; i++)            for (long long j = 0; j < b.len; j++)                c.s[i+j] += s[i] * b.s[j];        for (long long i = 0; i < c.len-1; i++)        {            c.s[i+1] += c.s[i] / 10;            c.s[i] %= 10;        }        c.clean();        return c;    }    /*高精度除以低精度*/    /*用到除法和取余的时候可能需要把全局的int替换成long long*/    bign operator / (long long b) const    {        assert(b > 0);        bign c;        c.len = len;        for (long long i = len-1, g = 0; i >= 0; --i)        {            long long x = 10*g+s[i];    //这里可能会超过int 故用long long            c.s[i] = x/b;                //这里可能会超过int            g = x-c.s[i]*b;                //这里可能会超过int        }        c.clean();        return c;    }    /*高精度对低精度取余*/    /*用到除法和取余的时候可能需要把全局的int替换成long long*/    bign operator % (long long b)    {        assert(b > 0);        bign d = b;        bign c = *this-*this/b*d;        return c;    }    bool operator < (const bign& b) const    {        if (len != b.len) return len < b.len;        for (long long i = len-1; i >= 0; i--)            if (s[i] != b.s[i]) return s[i] < b.s[i];        return false;    }    bool operator > (const bign& b) const    {        return b < *this;    }    bool operator <= (const bign& b)    {        return !(b > *this);    }    bool operator >= (const bign& b)    {        return !(b < *this);    }    bool operator == (const bign& b)    {        return !(b < *this) && !(*this < b);    }    bool operator != (const bign& b)    {        return (b < *this) || (*this < b);    }    bign operator += (const bign& b)    {        *this = *this + b;        return *this;    }};istream& operator >> (istream &in, bign& x){    string s;    in >> s;    x = s.c_str();    return in;}ostream& operator << (ostream &out, const bign& x){    out << x.str();    return out;}bign f[30 + 10];void init(){    f[0] = 1;    f[2] = 3;    bign four = 4;    for (int i = 4; i <= 30; i++)        f[i] = four * f[i - 2] - f[i - 4];}int main(){#ifdef LOCAL    freopen("in.txt", "r" , stdin);#endif    init();    int n;    while (cin >> n && n != -1)        cout << f[n] << endl;}