poj 2109Power of Cryptography

Power of Cryptography

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 16102		Accepted: 8119

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).

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<10¹⁰¹ and there exists an integer k, 1<=k<=10⁹ such that kⁿ = 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

Source

México and Central America 2004

大数相乘加二分

#include<stdio.h>
#include<string.h>
int n,lena,lenb;
int a[100000];
int b[100000];
int sum[100000];
char c[100000];
void dashu()
{
     memset(sum,0,sizeof(sum));//相当于一个临时变量，每次都得更新，初始化为0
     int i,j;
     for(i=0;i<lena;i++)//大数相乘算法
     for(j=0;j<lenb;j++)
     sum[i+j]+=a[i]*b[j];//a不断累乘，b不变
     for(i=0;i<lena*2;i++)
     {
          if(sum[i]>=10)
          {
               sum[i+1]+=sum[i]/10;
               sum[i]%=10;
          }
     }
     for(i=lena*2;i>=0;i--)//计算a数组的长度
     {
          if(sum[i]!=0)
          {
               lena=i+1;
               break;
          }
     }
     for(i=0;i<lena;i++)//把sum数组的值赋给a；
     a[i]=sum[i];
}
int judge(int mid)
{
     memset(a,0,sizeof(a));
     memset(b,0,sizeof(b));
  int i=0,j=0,len;
  while(mid!=0)//将其分离，用数组装
  {
       a[i++]=mid%10;

       b[j++]=a[i-1];
       mid/=10;
  }

  lena=i;
  lenb=i;
  for(i=0;i<n-1;i++)//进行n-1次乘法就行了
    dashu();
  if(lena>strlen(c))
  return 1;
  else if(lena<strlen(c))
  return -1;
  else//注意，当他们两的长度相同时，还要一个个比较
  {
       for(i=0;i<lena;i++)
       {
            if(c[i]-'0'!=a[lena-i-1])
            {
                 if(c[i]-'0'>a[lena-i-1])//该数比目标数小
                 return -1;
                 else
                 return 1;
            }
       }
  }
  return 0;//当返回0时说明这两个数相等
}
int main()
{

//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
     while(scanf("%d %s",&n,c)!=EOF)
     {
          int left=0;
          int right=100000000;
          int mid,ans=0;
          while(left<right)//k的范围为1到100000000二分就行了
          {
               mid=(left+right)/2;
              
               ans=judge(mid);//判断函数
               if(ans==1)//该数比目标数大，要变小，缩小范围
               right=mid;
               else if(ans==-1)//该数比目标数大，范围扩大
               left=mid+1;
               else if(ans==0)//找到，跳出
               break;
          }
          printf("%d
",mid);
     }
     return 0;
}

不过这题有一个一行的神级代码pow(p,1.0/n);

这样也能过，不过要用cout输出用printf的话就错了。