poj-3982(矩阵快速幂+大数模板)

问题描述:

数列A满足An = An-1 + An-2 + An-3, n >= 3 ,编写程序，给定A0, A1 和 A2, 计算A99

Input

输入包含多行数据 
每行数据包含3个整数A0, A1, A2 (0 <= A0, A1, A2 <= 32767) 
数据以EOF结束

Output

对于输入的每一行输出A99的值

Sample Input

1 1 1

Sample Output

69087442470169316923566147

题目题意:题目题意很简单，就是求A99.

题目分析:因为存在递推式的缘故，很自然的想到了矩阵快速幂来加速递推，但是这不是这道题目的关键，这道题目的范围，数据范围很大，导致没办法用普通数据范围来保存，所以大数的模板就很重要了。这道题目主要想分享一下这个大数模板.

代码如下:

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<string>using namespace std;const int MAXN =510;struct bign{    int len, s[MAXN];    bign ()    {        memset(s, 0, sizeof(s));        len = 1;    }    bign (int num) { *this = num; }    bign (const char *num) { *this = num; }    bign operator = (const int num)    {        char s[MAXN];        sprintf(s, "%d", num);        *this = s;        return *this;    }    bign operator = (const char *num)    {        for(int i = 0; num[i] == '0'; num++) ;  //去前导0        len = strlen(num);        for(int i = 0; i < len; i++) s[i] = num[len-i-1] - '0';        return *this;    }    bign operator + (const bign &b) const //+    {        bign c;        c.len = 0;        for(int i = 0, g = 0; g || i < max(len, b.len); i++)        {            int 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;    }    bign operator += (const bign &b)    {        *this = *this + b;        return *this;    }    void clean()    {        while(len > 1 && !s[len-1]) len--;    }    bign operator * (const bign &b) //*    {        bign c;        c.len = len + b.len;        for(int i = 0; i < len; i++)        {            for(int j = 0; j < b.len; j++)            {                c.s[i+j] += s[i] * b.s[j];            }        }        for(int i = 0; i < c.len; i++)        {            c.s[i+1] += c.s[i]/10;            c.s[i] %= 10;        }        c.clean();        return c;    }    bign operator *= (const bign &b)    {        *this = *this * b;        return *this;    }    bign operator - (const bign &b)    {        bign c;        c.len = 0;        for(int i = 0, g = 0; i < len; i++)        {            int 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;    }    bign operator -= (const bign &b)    {        *this = *this - b;        return *this;    }    bign operator / (const bign &b)    {        bign c, f = 0;        for(int i = len-1; i >= 0; i--)        {            f = f*10;            f.s[0] = s[i];            while(f >= b)            {                f -= b;                c.s[i]++;            }        }        c.len = len;        c.clean();        return c;    }    bign operator /= (const bign &b)    {        *this  = *this / b;        return *this;    }    bign operator % (const bign &b)    {        bign r = *this / b;        r = *this - r*b;        return r;    }    bign operator %= (const bign &b)    {        *this = *this % b;        return *this;    }    bool operator < (const bign &b)    {        if(len != b.len) return len < b.len;        for(int 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)    {        if(len != b.len) return len > b.len;        for(int 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)    {        return !(*this > b) && !(*this < b);    }    bool operator != (const bign &b)    {        return !(*this == b);    }    bool operator <= (const bign &b)    {        return *this < b || *this == b;    }    bool operator >= (const bign &b)    {        return *this > b || *this == b;    }    string str() const    {        string res = "";        for(int i = 0; i < len; i++) res = char(s[i]+'0') + res;        return res;    }};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;}//大数模板struct matrix//构建矩阵{    bign f[4][4];    matrix operator *( matrix &a) {        matrix res;        for (int i=1;i<=3;i++) {            for (int j=1;j<=3;j++) {                res.f[i][j]=0;                for (int k=1;k<=3;k++)                  res.f[i][j]+=f[i][k]*a.f[k][j];            }        }        return res;    }}a,b;int a0,a1,a2;void init(){    b.f[1][1]=b.f[1][2]=b.f[1][3]=1;    b.f[2][1]=1;    b.f[3][2]=1;    a.f[1][1]=a2,a.f[2][1]=a1,a.f[3][1]=a0;}matrix fast_pow(matrix base,int k)//快速幂{    matrix ans;    for (int i=1;i<=3;i++)        ans.f[i][i]=1;    while (k) {        if (k&1)          ans=ans*base;        base=base*base;        k>>=1;    }    return ans;}int main(){    while (scanf("%d%d%d",&a0,&a1,&a2)!=EOF) {        init();        matrix cur,ans;        cur=b;        cur=fast_pow(cur,97);        ans=cur*a;        cout<<ans.f[1][1]<<endl;    }    return 0;}