HDU1041 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;
}