1169: 简短而优美的题目

题目描述：

    在lfx高尚又有趣的灵魂的影响下，我想出一个简短而优美的题目。

    This is too simple a question.

    Given an integer N, find the value of this formula C (n, 0)+C (n, 1)+C (n, 2)+...+C (n, n-1)+C (n, n)，And 1<=n<=1000

Among them, the definition of C (n, m) combination: from n different elements, m (m <= n) elements are selected and grouped together, which is called a combination of M elements from n different elements; from n different elements, the number of all combinations of M (m <= n) elements is called the number of combinations of M elements from n different elements. Represented by the symbol C (n, m).

思路：

英文题面就是让大家熟悉英文题面，因为入门后就没什么中文题了。

思路：

C (n, 0)+C (n, 1)+C (n, 2)+...+C (n, n-1)+C (n, n) = 2^n， 高中的数学公式。

然后问题转化为求2^n，n很大，显然用int or long long 是没法solve的。

那么我们这里给出两个方法，

一个是double抖机灵。

double 的数据类型取值范围如下：

1、负值取值范围为 -1.79769313486231570E+308 到 -4.94065645841246544E-324；

2、正值取值范围为 4.94065645841246544E-324 到 1.79769313486231570E+308

但是当数值真的很大的时候，比如1e100，double不会爆，但是会精度丢失，即和真正的数值有偏差，所以一半很大的数也不用double来表示。

但是这题是求2的N次方，应该知道的是，计算机表示一个数使用二进制，那么2的N次无论多少位都是只有一位数值为1，其他全为0。

那么double就不会精度丢失，（因为就1位为1）

所以这一题就可以用double直接AC。

见代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define rt return
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d
",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define db(x) cout<<"== [ "<<x<<" ] =="<<endl;
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
inline void getInt(int* p);
const int maxn=1000010;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
int n;
int main()
{
//    freopen("D:\common_text\code_stream\in.txt","r",stdin);
//    freopen("D:\common_text\code_stream\out.txt","w",stdout);
    scanf("%d",&n);
    printf("%.0f
",pow(2,n));
    return 0;
}
 
inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '
');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    }
    else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}

第二种方法是直接使用c++大数模板

见代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define rt return
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d
",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define db(x) cout<<"== [ "<<x<<" ] =="<<endl;
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
inline void getInt(int* p);
const int maxn=1000010;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
const int MAXN=2500;
struct bign
{
    int len, s[MAXN];
    bign ()
    {
        memset(s, 0, sizeof(s));
        len = 1;
    }
    bign (int num) { *this = num; }
    bign (const char *num) { *this = num; }
    bign operator = (const int num)
    {
        char s[MAXN];
        sprintf(s, "%d", num);
        *this = s;
        return *this;
    }
    bign operator = (const char *num)
    {
        for(int i = 0; num[i] == '0'; num++) ;  //去前导0
        len = strlen(num);
        for(int i = 0; i < len; i++) s[i] = num[len-i-1] - '0';
        return *this;
    }
    bign operator + (const bign &b) const //+
    {
        bign c;
        c.len = 0;
        for(int i = 0, g = 0; g || i < max(len, b.len); i++)
        {
            int x = g;
            if(i < len) x += s[i];
            if(i < b.len) x += b.s[i];
            c.s[c.len++] = x % 10;
            g = x / 10;
        }
        return c;
    }
    bign operator += (const bign &b)
    {
        *this = *this + b;
        return *this;
    }
    void clean()
    {
        while(len > 1 && !s[len-1]) len--;
    }
    bign operator * (const bign &b) //*
    {
        bign c;
        c.len = len + b.len;
        for(int i = 0; i < len; i++)
        {
            for(int j = 0; j < b.len; j++)
            {
                c.s[i+j] += s[i] * b.s[j];
            }
        }
        for(int i = 0; i < c.len; i++)
        {
            c.s[i+1] += c.s[i]/10;
            c.s[i] %= 10;
        }
        c.clean();
        return c;
    }
    bign operator *= (const bign &b)
    {
        *this = *this * b;
        return *this;
    }
    bign operator - (const bign &b)
    {
        bign c;
        c.len = 0;
        for(int i = 0, g = 0; i < len; i++)
        {
            int x = s[i] - g;
            if(i < b.len) x -= b.s[i];
            if(x >= 0) g = 0;
            else
            {
                g = 1;
                x += 10;
            }
            c.s[c.len++] = x;
        }
        c.clean();
        return c;
    }
    bign operator -= (const bign &b)
    {
        *this = *this - b;
        return *this;
    }
    bign operator / (const bign &b)
    {
        bign c, f = 0;
        for(int i = len-1; i >= 0; i--)
        {
            f = f*10;
            f.s[0] = s[i];
            while(f >= b)
            {
                f -= b;
                c.s[i]++;
            }
        }
        c.len = len;
        c.clean();
        return c;
    }
    bign operator /= (const bign &b)
    {
        *this  = *this / b;
        return *this;
    }
    bign operator % (const bign &b)
    {
        bign r = *this / b;
        r = *this - r*b;
        return r;
    }
    bign operator %= (const bign &b)
    {
        *this = *this % b;
        return *this;
    }
    bool operator < (const bign &b)
    {
        if(len != b.len) return len < b.len;
        for(int i = len-1; i >= 0; i--)
        {
            if(s[i] != b.s[i]) return s[i] < b.s[i];
        }
        return false;
    }
    bool operator > (const bign &b)
    {
        if(len != b.len) return len > b.len;
        for(int i = len-1; i >= 0; i--)
        {
            if(s[i] != b.s[i]) return s[i] > b.s[i];
        }
        return false;
    }
    bool operator == (const bign &b)
    {
        return !(*this > b) && !(*this < b);
    }
    bool operator != (const bign &b)
    {
        return !(*this == b);
    }
    bool operator <= (const bign &b)
    {
        return *this < b || *this == b;
    }
    bool operator >= (const bign &b)
    {
        return *this > b || *this == b;
    }
    string str() const
    {
        string res = "";
        for(int i = 0; i < len; i++) res = char(s[i]+'0') + res;
        return res;
    }
};
istream& operator >> (istream &in, bign &x)
{
    string s;
    in >> s;
    x = s.c_str();
    return in;
}
ostream& operator << (ostream &out, const bign &x)
{
    if (x.str()=="") out<<0;
    else out << x.str();
    return out;
}

int main()
{
    int n;
    cin>>n;
    bign ans=1;
    repd(i,1,n)
    {
        ans=ans*2;
    }
    cout<<ans<<endl;
    return 0;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '
');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    }
    else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}

当然，JAVA和python均可以解决大数题。