【二分+高精度快速幂】poj2109
Power of Cryptography
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 17140 | Accepted: 8660 |
Description
Current work in cryptography involves (among other things) large prime numbers and computing powers of numbers among these primes. Work in this area has resulted in the practical use of results from number theory and other branches of mathematics once considered to be only of theoretical interest.
This problem involves the efficient computation of integer roots of numbers.
Given an integer n>=1 and an integer p>= 1 you have to write a program that determines the n th positive root of p. In this problem, given such integers n and p, p will always be of the form k to the n th. power, for an integer k (this integer is what your program must find).
This problem involves the efficient computation of integer roots of numbers.
Given an integer n>=1 and an integer p>= 1 you have to write a program that determines the n th positive root of p. In this problem, given such integers n and p, p will always be of the form k to the n th. power, for an integer k (this integer is what your program must find).
Input
The input consists of a sequence of integer pairs n and p with each integer on a line by itself. For all such pairs 1<=n<= 200, 1<=p<10 101 and there exists an integer k, 1<=k<=10 9 such that k n = p.
Output
For each integer pair n and p the value k should be printed, i.e., the number k such that k n =p.
Sample Input
2 16 3 27 7 4357186184021382204544
Sample Output
4 3 1234
题目大意：给出n和p 求出k使k^n=p
看数据范围就知道是要用二分加快速幂，还要高精度……
这个题从12月11号开始做的，但是那一天以后就停竞赛期末复习了，于是没有搞出来的程序一直扔在那儿，今天过生日，奖励自己刷两个题，于是终于过了。
其实一点也不难，不过最开始有好多bug啊！！现在写个高精度各种错，退化啊退化……
还有，二分貌似写得不很科学啊啊啊……
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
long n;
char s[200];
long p[120];
long l[120], r[120], mid[120] ,re[200];
long t[200], y[200];
bool comp1(long *l, long *r)
{
if (l[0] < r[0]) return 1;
if (l[0] > r[0]) return 0;
for (long i=l[0]; i>0; i--)
{
if (l[i] < r[i]) return 1;
if (l[i] > r[i]) return 0;
}
return 0;
}
bool comp_deng(long *l, long *r)
{
if (l[0] != r[0]) return 0;
for (long i=l[0]; i>0; i--)
{
if (l[i] != r[i]) return 0;
}
return 1;
}
void sum_div2()
{
memset(mid, 0, sizeof(mid));
long len = r[0];
for (long i=1; i<=len; i++)
{
mid[i] += l[i] + r[i];
mid[i + 1] = mid[i] / 10;
mid[i] = mid[i] % 10;
}
if (mid[len + 1] > 0) len++;
for (long i=len; i>=1; i--)
{
mid[i-1] += mid[i] % 2 * 10;
mid[i] = mid[i] /2;
}
while (mid[len] <= 0) len--;
mid[0] = len;
}
void cheng(long *a, long *b)
{
long tmp[400];
memset(tmp, 0, sizeof(tmp));
for (long i=1; i<=a[0]; i++)
for (long j=1; j<=b[0]; j++)
{
tmp[i + j -1] += a[i] * b[j];
tmp[i + j] += tmp[i + j -1] / 10;
tmp[i + j -1] %= 10;
}
long len = a[0] + b[0];
while (tmp[ len ] == 0) len--;
if (len >= 200) { len = 200; tmp[200] = 9; }
tmp[0] = len;
for (long i=0; i<=tmp[0]; i++)
{
a[i] = tmp[i];
}
}
void power(long n)
{
memset(t, 0, sizeof(t));
t[0] = 1;
t[1] = 1;
for (long i=0; i<=mid[0]; i++)
{
y[i] = mid[i];
}
while (n != 0)
{
if ((n & 1) == 1)
{
cheng(t, y);
}
cheng(y, y);
n = n >> 1;
}
for (long i=0; i<=t[0]; i++)
{
re[i] = t[i];
}
}
void add1()
{
for (long i=0; i<=mid[0]; i++)
{
l[i] = mid[i];
}
l[1] += 1;
long i = 1;
while (l[i] >= 10)
{
l[i + 1] += l[i] / 10;
l[i] = l[i] % 10;
i++;
}
if (l[l[0] + 1] > 0) l[0]++;
}
void plus1()
{
for (long i=0; i<=mid[0]; i++)
{
r[i] = mid[i];
}
r[1] -= 1;
long i = 1;
while (r[i] < 0)
{
r[i + 1] -= 1;
r[i] += 10;
i--;
}
if (l[l[0]] == 0) l[0]--;
}
void merge(long *a)
{
while (comp1(l, r))
{
sum_div2();
power(n);
if (comp_deng(re, p))
{
for (long i=0; i<=mid[0]; i++)
{
a[i] = mid[i];
}
return;
}
if (comp1(re, p)) add1();
else plus1();
}
for (long i=0; i<=l[0]; i++)
{
a[i] = l[i];
}
return;
}
int main()
{
// freopen("2109.in", "r", stdin);
while (cin >> n >> s)
{
memset(p, 0, sizeof(p));
memset(r, 0, sizeof(r));
long len = strlen(s);
for (long i=0; i<len; i++)
{
p[len - i] = s[i] - '0';
r[len - i] = s[i] - '0';
}
p[0] = len;
r[0] = len;
memset(l, 0, sizeof(l));
l[0] = 1; l[1] = 1;
merge(re);
for (long i=re[0]; i>0; i--)
{
cout<<re[i];
}
cout<<endl;
}
return 0;
}