BZOJ 2656 ZJOI 2012 数列(sequence) 高精度+记忆化搜索

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题目大意：定义个一序列，f[i] = f[i / 2] (i % 2 == 0);f[i] = f[i / 2] + f[i / 2 + 1] (i % 2 == 1);求这个数列的第m项（m <= 10 ^ 100）

思路：数据范围高精度没跑了。记得之前做过这个题的弱化版，似乎是没有高精度的记忆化搜索，这个题就是加个高精度。

CODE：

#include <map>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define MAX 210#define MO 10using namespace std;#define max(a,b) ((a) > (b) ? (a):(b)) class BigInt{private:    int num[MAX],len;public:    BigInt(int _ = 0) {        memset(num,0,sizeof(num));        num[len = 1] = _;    }    void Read() {        static char s[MAX];        scanf("%s",s);        len = strlen(s);        for(int i = 1; i <= len; ++i)            num[i] = s[len - i] - '0';    }    friend ostream &operator <<(ostream &os,const BigInt &a);    bool operator <(const BigInt &a)const {        if(len == a.len) {            for(int i = len; i; --i)                if(num[i] != a.num[i])                    return num[i] < a.num[i];        }        return len < a.len;    }    bool Odd()const {        return num[1]&1;    }    BigInt operator +(const BigInt &a)const {        BigInt re;        re.len = max(len,a.len);        int temp = 0;        for(int i = 1; i <= re.len; ++i) {            re.num[i] = temp + num[i] + a.num[i];            temp = re.num[i] / MO;            re.num[i] %= MO;        }        if(temp)    re.num[++re.len] = temp;        return re;    }    BigInt operator /(int c)const {        BigInt re;        re.len = len;        int temp = 0;        for(int i = len; i; --i) {            re.num[i] = (num[i] + temp) / 2;            temp = (num[i] + temp) % 2 * MO;        }        while(!re.num[re.len])  --re.len;        return re;    }}temp; ostream &operator <<(ostream &os,const BigInt &a){    os << a.num[a.len];    for(int i = a.len - 1; i; --i)        os << a.num[i];    return os;} map<BigInt,BigInt> G; int T; BigInt &MemorialSearch(const BigInt &a){    if(G.find(a) != G.end())    return G[a];    if(a.Odd())        G[a] = MemorialSearch(a / 2) + MemorialSearch(a / 2 + 1);    else    G[a] = MemorialSearch(a / 2);    return G[a];} int main(){    G[0] = BigInt(0);    G[1] = BigInt(1);    for(cin >> T; T--;) {        temp.Read();        cout << MemorialSearch(temp) << endl;    }    return 0;}

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