POJ 2109 Power of Cryptography【高精度+二分 Or double水过~~】

题目链接：

http://poj.org/problem?id=2109

参考：

http://blog.csdn.net/code_pang/article/details/8263971

题意：

给定 n,p ，求 k 使得 kn=p ( 1≤n≤200 , 1≤p<10101 , 1≤k≤109 )

分析：

高精度+二分~~
k的位数为p的位数除以n的向上取整，这样知道k的位数便可以在范围内二分了~注意这里的答案是向下取整~~

代码：

#include<cstring>
#include<cstdio>
#include<cmath>
#include<cstdlib>
using namespace std;
#define MAXN 9999
#define MAXSIZE 1000
#define DLEN 4
const int maxn = 205;
char p[maxn];
class BigNum
 {
    private:
        int a[500];
        int len;
    public:
        BigNum(){
            len = 1;
            memset(a, 0 ,sizeof(a));
        }
        BigNum(const int);
        BigNum(const char*);
        BigNum(const BigNum &);
        BigNum &operator = (const BigNum &);

        BigNum operator +(const BigNum &)const ;
        BigNum operator*(const BigNum &)const ;
        BigNum operator^(const int &)const ;
        bool operator >(const int &)const;
        bool operator >(const BigNum &T)const;

        void print();
 };
 BigNum::BigNum(const int b)
 {
     int c, d = b;
     len = 0;
     while(d > MAXN){
        c = d % (MAXN + 1);
        d = d / (MAXN + 1);
        a[len++] = c;
     }
     a[len++] = d;
 }
BigNum::BigNum(const char*s)
{
    int t, k, index;
    memset(a,0,sizeof(a));
    int l = strlen(s);
    len = l / DLEN;
    if(l % DLEN)
        len++;
    index=0;
    for(int i = l - 1; i >= 0; i -= DLEN){
        t = 0;
        k = i - DLEN + 1;
        if(k<0) k=0;
        for(int j = k; j <= i; j++)
            t = t * 10 + s[j] - '0';
        a[index++] = t;
    }
}
BigNum::BigNum(const BigNum & T) : len(T.len)
{
    memset(a, 0, sizeof(a));
    for(int i = 0 ; i < len ; i++)
        a[i] = T.a[i];
}
BigNum & BigNum::operator=(const BigNum & n)
{
    len = n.len;
    memset(a, 0, sizeof(a));
    for(int i = 0 ; i < len ; i++)
        a[i] = n.a[i];
    return *this;
}
BigNum BigNum::operator+(const BigNum & T) const
{
    BigNum t(*this);
    int big;      //位数
    big = T.len > len ? T.len : len;
    for(int i = 0; i < big; i++){
        t.a[i] +=T.a[i];
        if(t.a[i] > MAXN){
            t.a[i + 1]++;
            t.a[i] -=MAXN+1;
        }
    }
    if(t.a[big] != 0)  t.len = big + 1;
    else t.len = big;
    return t;
}
BigNum BigNum::operator*(const BigNum & T) const
{
    BigNum ret;
    int up;
    int temp,temp1;
    int i, j;
    for(i = 0 ; i < len ; i++){
        up = 0;
        for(j = 0 ; j < T.len ; j++){
            temp = a[i] * T.a[j] + ret.a[i + j] + up;
            if(temp > MAXN){
                temp1 = temp - temp / (MAXN + 1) * (MAXN + 1);
                up = temp / (MAXN + 1);
                ret.a[i + j] = temp1;
            }
            else{
                up = 0;
                ret.a[i + j] = temp;
            }
        }
        if(up != 0)
            ret.a[i + j] = up;
    }
    ret.len = i + j;
    while(ret.a[ret.len - 1] == 0 && ret.len > 1)
        ret.len--;
    return ret;
}
BigNum BigNum::operator^(const int & n) const
{
    BigNum t,ret(1);
    if(n < 0)  exit(-1);
    if(n == 0)  return 1;
    if(n == 1) return *this;
    int m = n;
    int i;
    while(m > 1){
        t = *this;
        for(i = 1; i<<1<=m; i <<= 1)
            t = t * t;
        m -= i;
        ret = ret * t;
        if(m == 1) ret=ret*(*this);
    }
    return ret;
}
bool BigNum::operator>(const BigNum & T) const
{
    int ln;
    if(len > T.len)
        return true;
    else if(len == T.len){
        ln = len - 1;
        while(a[ln] == T.a[ln] && ln >= 0)
            ln--;
        if(ln >= 0 && a[ln] > T.a[ln])
            return true;
        else
            return false;
    }
    else
        return false;
}
bool BigNum::operator >(const int & t) const
{
    BigNum b(t);
    return *this>b;
}
void BigNum::print()
{
    printf("%d",a[len-1]);
    for(int i = len - 2 ; i >= 0 ; i--)  printf("%04d",a[i]);
    printf("\n");
}
int main(void)
{
    int n, len, MIN, MAX, MID;
    while(~scanf("%d%s", &n, &p)){
        len = (int)ceil((double)strlen(p) / n);
        MIN = 1, MAX = 9;
        for(int i = 0; i < len - 1; i++){
            MAX *= 10;
            MAX += 9;
            MIN *= 10;
        }
        while(MIN < MAX){//[]
            MID = (MIN + MAX) / 2;
            if(BigNum(p) > (BigNum(MID) ^ n)) MIN = MID +1;
            else if((BigNum(MID) ^ n) > BigNum(p)) MAX = MID - 1;
            else break;
        }
        if(MAX == MIN)  MID = MIN;
        if((BigNum(MID) ^ n) > BigNum(p)) MID--;
         printf("%d\n",MID);
    }
    return 0;
}

代码：

double类型能表示 10−307到10308 , 足够这个题用。
而double超过16位后面都变成0，这样正好满足向下取整。
12337=4332529576639313702577
12347=4357186184021382204544
12357=4381962969567270546875
值不同的地方（从高到低第三位）没有超过double的精度，所以不会导致错误答案~

#include<iostream>
#include<cmath>
using namespace std;
int main (void)
{
    double n, m;
    while(cin>>n>>m){
        cout<<pow(m, 1/n)<<endl;
    }
    return 0;
}

//或者

#include<iostream>
#include<cmath>
using namespace std;
int main (void)
{
    double n, p;
    while(cin>>n>>p){
         cout<< exp(log(p)/n) <<endl;
    }
    return 0;
}
//pow明显更快，只是想说明有时候取对数也是个不错的方法~