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~