今日SGU 5.29

sgu 299

题意:给你n个线段,然后问你能不能选出其中三个组成一个三角形,数字很大

收获:另一个大整数模板

#include<bits/stdc++.h>
#define de(x) cout<<#x<<"="<<x<<endl;
#define dd(x) cout<<#x<<"="<<x<<" ";
#define rep(i,a,b) for(int i=a;i<(b);++i)
#define repd(i,a,b) for(int i=a;i>=(b);--i)
#define repp(i,a,b,t) for(int i=a;i<(b);i+=t)
#define ll long long
#define mt(a,b) memset(a,b,sizeof(a))
#define fi first
#define se second
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f3f3f
#define pii pair<int,int>
#define pdd pair<double,double>
#define pdi pair<double,int>
#define mp(u,v) make_pair(u,v)
#define sz(a) (int)a.size()
#define ull unsigned long long
#define ll long long
#define pb push_back
#define PI acos(-1.0)
#define qc std::ios::sync_with_stdio(false)
#define db double
#define all(a) a.begin(),a.end()
const int mod = 1e9+7;
const int maxn = 1e3+6;
const double eps = 1e-6;
using namespace std;
class bign  
{  
public:  
    int len, s[maxn];//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[maxn] = "";  
    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[maxn];  
    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;  
}
void bign::clean()  
{  
    if (len == 0) len++;  
    while (len > 1 && s[len - 1] == '\0')  
        len--;  
}
bign::~bign()  
{  
}
bign a[maxn];
int main(){
    int n;
    scanf("%d",&n);
    rep(i,1,n+1) cin>>a[i];
    sort(a+1,a+n+1);
    rep(i,3,n+1) if(a[i] < a[i-1] + a[i-2]) {
        cout<<a[i]<<" "<<a[i-1]<<" "<<a[i-2]<<endl;
        return 0;
    }
    puts("0 0 0");
    return 0;
}
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转载自www.cnblogs.com/chinacwj/p/9107414.html