OpenJ_POJ C16D Extracurricular Sports 找规律、大整数类

A - 莽撞人

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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Description

As we all know, to pass PE exams, students need to do extracurricular sports, especially jogging. As the result, the Jogging Association 

in the university is very popular.

There are N students in the Jogging Association. Some run faster while the others run slower. The time it takes for a student to run 

exactly one lap is called his/her ''lap time'', which is always a positive integer. The lap times of students are distinct from each other. However, running too slow is not allowed for students of the Jogging Association, so each lap time is at most 100 digits.

One day, they make an appointment to jog together. They begin at the same time as well as the same starting point, jog along the 

circular runway at their own steady speed round and round, not stop until the first time when all of them meet again at the starting point. When 

they stop, they accidently find that the time they have spent is precisely equal to the time they need if everyone runs a lap one by one, 

as a relay race.

Now we are curious about the lap times of all the students. Can you guess out any possible solution?

Input

The first line contains an integer T (1 ≤ T ≤ 30), indicating the number of test cases. 

For each test case: 

A line contains an integer N (2 ≤ N ≤ 200), indicating the number of students.

Output

For each test case: 

If the solution does not exist, output -1 on a single line; 

Otherwise, output one solution with n different integers t1,t2,...,tn, one per line, where ti (1 ≤ ti < 10^100) indicates the lap time of the   i-th student. If there are several solutions, output any one.

Sample Input

1
3

Sample Output

10
20
30

Hint

For the three students in the Sample Output, they will stop at time 60, which is precisely equal to the time they need if they run as a relay race.

Source

UESTC 2016 Summer Training #14 Div.2

OpenJ_POJ C16D

My Solution

做这种题感觉比较碰运气(水平不够， 所以比较看运气了)

这个是找规律的， 写几组然后看看

所有数的最大公倍数 == 所有数的和

1 2 3

1 2 6 9

1 2 6 18 27

这里前 n - 1个数是 1 2 6 18 这样 以3为公比增加的， 然后每个n 对应的最后一个数 是 3^n次

然后套一个大整数类的版就好啦

这个大整数类的版很棒的

#include <iostream>
#include <cstdio>
#include <string>
#include <iosfwd>
#include <cmath>
#include <cstring>
#include <stdlib.h>

#define MAX_L 2005 //最大长度，可以修改
using namespace std;

class bign
{
public:
    int len, s[MAX_L];//数的长度，记录数组
//构造函数
    bign();
    bign(const char*);
    bign(int);
    bool sign;//符号 1正数 0负数
    string toStr() const;//转化为字符串，主要是便于输出
    friend istream& operator>>(istream &,bign &);//重载输入流
    friend ostream& operator<<(ostream &,bign &);//重载输出流
//重载复制
    bign operator=(const char*);
    bign operator=(int);
    bign operator=(const string);
//重载各种比较
    bool operator>(const bign &) const;
    bool operator>=(const bign &) const;
    bool operator<(const bign &) const;
    bool operator<=(const bign &) const;
    bool operator==(const bign &) const;
    bool operator!=(const bign &) const;
//重载四则运算
    bign operator+(const bign &) const;
    bign operator++();
    bign operator++(int);
    bign operator+=(const bign&);
    bign operator-(const bign &) const;
    bign operator--();
    bign operator--(int);
    bign operator-=(const bign&);
    bign operator*(const bign &)const;
    bign operator*(const int num)const;
    bign operator*=(const bign&);
    bign operator/(const bign&)const;
    bign operator/=(const bign&);
//四则运算的衍生运算
    bign operator%(const bign&)const;//取模（余数）
    bign factorial()const;//阶乘
    bign Sqrt()const;//整数开根（向下取整）
    bign pow(const bign&)const;//次方
//一些乱乱的函数
    void clean();
    ~bign();
};
#define max(a,b) a>b ? a : b
#define min(a,b) a<b ? a : b

bign::bign()
{
    memset(s, 0, sizeof(s));
    len = 1;
    sign = 1;
}

bign::bign(const char *num)
{
    *this = num;
}

bign::bign(int num)
{
    *this = num;
}

string bign::toStr() const
{
    string res;
    res = "";
    for (int i = 0; i < len; i++)
        res = (char)(s[i] + '0') + res;
    if (res == "")
        res = "0";
    if (!sign&&res != "0")
        res = "-" + res;
    return res;
}

istream &operator>>(istream &in, bign &num)
{
    string str;
    in>>str;
    num=str;
    return in;
}

ostream &operator<<(ostream &out, bign &num)
{
    out<<num.toStr();
    return out;
}

bign bign::operator=(const char *num)
{
    memset(s, 0, sizeof(s));
    char a[MAX_L] = "";
    if (num[0] != '-')
        strcpy(a, num);
    else
        for (int i = 1; i < strlen(num); i++)
            a[i - 1] = num[i];
    sign = !(num[0] == '-');
    len = strlen(a);
    for (int i = 0; i < strlen(a); i++)
        s[i] = a[len - i - 1] - 48;
    return *this;
}

bign bign::operator=(int num)
{
    if (num < 0)
        sign = 0, num = -num;
    else
        sign = 1;
    char temp[MAX_L];
    sprintf(temp, "%d", num);
    *this = temp;
    return *this;
}

bign bign::operator=(const string num)
{
    const char *tmp;
    tmp = num.c_str();
    *this = tmp;
    return *this;
}

bool bign::operator<(const bign &num) const
{
    if (sign^num.sign)
        return num.sign;
    if (len != num.len)
        return len < num.len;
    for (int i = len - 1; i >= 0; i--)
        if (s[i] != num.s[i])
            return sign ? (s[i] < num.s[i]) : (!(s[i] < num.s[i]));
    return !sign;
}

bool bign::operator>(const bign&num)const
{
    return num < *this;
}

bool bign::operator<=(const bign&num)const
{
    return !(*this>num);
}

bool bign::operator>=(const bign&num)const
{
    return !(*this<num);
}

bool bign::operator!=(const bign&num)const
{
    return *this > num || *this < num;
}

bool bign::operator==(const bign&num)const
{
    return !(num != *this);
}

bign bign::operator+(const bign &num) const
{
    if (sign^num.sign)
    {
        bign tmp = sign ? num : *this;
        tmp.sign = 1;
        return sign ? *this - tmp : num - tmp;
    }
    bign result;
    result.len = 0;
    int temp = 0;
    for (int i = 0; temp || i < (max(len, num.len)); i++)
    {
        int t = s[i] + num.s[i] + temp;
        result.s[result.len++] = t % 10;
        temp = t / 10;
    }
    result.sign = sign;
    return result;
}

bign bign::operator++()
{
    *this = *this + 1;
    return *this;
}

bign bign::operator++(int)
{
    bign old = *this;
    ++(*this);
    return old;
}

bign bign::operator+=(const bign &num)
{
    *this = *this + num;
    return *this;
}

bign bign::operator-(const bign &num) const
{
    bign b=num,a=*this;
    if (!num.sign && !sign)
    {
        b.sign=1;
        a.sign=1;
        return b-a;
    }
    if (!b.sign)
    {
        b.sign=1;
        return a+b;
    }
    if (!a.sign)
    {
        a.sign=1;
        b=bign(0)-(a+b);
        return b;
    }
    if (a<b)
    {
        bign c=(b-a);
        c.sign=false;
        return c;
    }
    bign result;
    result.len = 0;
    for (int i = 0, g = 0; i < a.len; i++)
    {
        int x = a.s[i] - g;
        if (i < b.len) x -= b.s[i];
        if (x >= 0) g = 0;
        else
        {
            g = 1;
            x += 10;
        }
        result.s[result.len++] = x;
    }
    result.clean();
    return result;
}

bign bign::operator * (const bign &num)const
{
    bign result;
    result.len = len + num.len;

    for (int i = 0; i < len; i++)
        for (int j = 0; j < num.len; j++)
            result.s[i + j] += s[i] * num.s[j];

    for (int i = 0; i < result.len; i++)
    {
        result.s[i + 1] += result.s[i] / 10;
        result.s[i] %= 10;
    }
    result.clean();
    result.sign = !(sign^num.sign);
    return result;
}

bign bign::operator*(const int num)const
{
    bign x = num;
    bign z = *this;
    return x*z;
}
bign bign::operator*=(const bign&num)
{
    *this = *this * num;
    return *this;
}

bign bign::operator /(const bign&num)const
{
    bign ans;
    ans.len = len - num.len + 1;
    if (ans.len < 0)
    {
        ans.len = 1;
        return ans;
    }

    bign divisor = *this, divid = num;
    divisor.sign = divid.sign = 1;
    int k = ans.len - 1;
    int j = len - 1;
    while (k >= 0)
    {
        while (divisor.s[j] == 0) j--;
        if (k > j) k = j;
        char z[MAX_L];
        memset(z, 0, sizeof(z));
        for (int i = j; i >= k; i--)
            z[j - i] = divisor.s[i] + '0';
        bign dividend = z;
        if (dividend < divid) { k--; continue; }
        int key = 0;
        while (divid*key <= dividend) key++;
        key--;
        ans.s[k] = key;
        bign temp = divid*key;
        for (int i = 0; i < k; i++)
            temp = temp * 10;
        divisor = divisor - temp;
        k--;
    }
    ans.clean();
    ans.sign = !(sign^num.sign);
    return ans;
}

bign bign::operator/=(const bign&num)
{
    *this = *this / num;
    return *this;
}

bign bign::operator%(const bign& num)const
{
    bign a = *this, b = num;
    a.sign = b.sign = 1;
    bign result, temp = a / b*b;
    result = a - temp;
    result.sign = sign;
    return result;
}

bign bign::pow(const bign& num)const
{
    bign result = 1;
    for (bign i = 0; i < num; i++)
        result = result*(*this);
    return result;
}

bign bign::factorial()const
{
    bign result = 1;
    for (bign i = 1; i <= *this; i++)
        result *= i;
    return result;
}

void bign::clean()
{
    if (len == 0) len++;
    while (len > 1 && s[len - 1] == '\0')
        len--;
}

bign bign::Sqrt()const
{
    if(*this<0)return -1;
    if(*this<=1)return *this;
    bign l=0,r=*this,mid;
    while(r-l>1)
    {
        mid=(l+r)/2;
        if(mid*mid>*this)
            r=mid;
        else
            l=mid;
    }
    return l;
}

bign::~bign()
{
}


const int maxn = 2*1e2 + 8;
bign val[maxn], ans[maxn];

int main()
{
    #ifdef LOCAL
    freopen("a.txt", "r", stdin);
    //freopen("b.txt", "w", stdout);
    #endif // LOCAL
    val[1] = 1;
    val[2] = 2;
    val[3] = 6;
    for(int i = 4; i < maxn; i++){
        val[i] = val[i-1] * 3;
    }
    ans[3] = 3;
    for(int i = 4; i < maxn; i++){
        ans[i] = ans[i-1] * 3;
    }


    int T, n;
    scanf("%d", &T);
    while(T--){
        scanf("%d", &n);
        if(n < 3) printf("-1\n");
        else{
            for(int i = 1; i < n; i++){
                cout<<val[i];
                printf("\n");
            }
            cout<<ans[n]<<endl;
        }
    }
    return 0;
}

  Thank you!

                                                                                                                                               ------from ProLights