HDU-1041 Computer Transformation 大数

这题打表找下规律就可以了，定义一个变量来表示增量，那么这个变量的格律就是 add = (add ± 1) * 2

手写了大数的类，幸好只有-1这个值，这个类是没定义减法运算的。

代码如下：

#include <cstdlib>

#include <cstdio>

#include <cstring>

#include <cmath>

#include <algorithm>

using namespace std;



typedef long long int Int64;



Int64 rec[65];



int N;



struct BigInteger

{

    char x[1005];

    BigInteger ()

    {

        memset(x, 0, sizeof (x));    

    }

    BigInteger operator + (BigInteger b);

    BigInteger operator * (int m);

    void print();

}ret[1005], Add;



BigInteger BigInteger :: operator + (BigInteger b)

{

    BigInteger ans;

    int lena = 0, lenb = 0, len;

    for (int i = 1000; i >= 0; --i) {

        if (x[i] && !lena) lena = i;

        if (b.x[i] && !lenb) lenb = i;

        if (lena && lenb) break;

    }

    len = max(lena, lenb);

    for (int i = 0; i <= len; ++i) {

        ans.x[i] += x[i] + b.x[i];

        if (ans.x[i] >= 10) {

            ans.x[i+1] += ans.x[i] / 10;

            ans.x[i] %= 10;

        }

    }

    return ans;

}



BigInteger BigInteger :: operator * (int m)

{

    BigInteger ans;

    int len = 0;

    for (int i = 1000; i >= 0; --i) {

        if (x[i] && !len) len = i;

        if (len) break;

    }

    for (int i = 0; i <= len; ++i) {

        ans.x[i] += x[i] * m;

        if (ans.x[i] >= 10) {

            ans.x[i+1] += ans.x[i] / 10;

            ans.x[i] %= 10;

        }    

    }

    return ans;

}



void BigInteger :: print()

{

    int len = 0;

    for (int i = 1000; i >= 0; --i) {

        if (x[i] && !len) len = i;

        if (len) break;

    } 

    for (int i = len; i >= 0; --i) {

        printf("%d", x[i]);    

    }

    puts("");

}



BigInteger valueof(Int64 obj)

{

    BigInteger ans;

    int i = 0;

    while (obj) {

        ans.x[i] = obj % 10;

        obj /= 10;

        ++i;

    }

    return ans;

}



int main()

{

    ret[1] = valueof(0), ret[2] = ret[3] = valueof(1); 

    Add = valueof(0);

    for (int i = 4; i <= 1000; ++i) { 

        Add = (Add + valueof(i & 1 ? -1 : 1)) * 2; 

        ret[i] = ret[i-1] + Add;

    }

    while (scanf("%d", &N) == 1) {

        ret[N].print();

    }

    return 0;

}