NYOJ773 开放数

题目：NYOJ773

描述

现在给你两个数 n 和 p ，让你求出 p 的开 n 次方。

输入

每组数据包含两个数n和p。当n和p都为0时表示输入结束。(1<=n<=200,1<=p<=10^101)

输出

对于每个输出对用输出开方后的结果k（结果小于10^9）。

样例输入

2 16
3 27
7 4357186184021382204544
0 0

样例输出

4
3
1234

自己写的二分求根超时，居然用pow可以过，我也是醉了，可能算法有bug，求纠正。

//er fen fa
#include<iostream>
#include<string.h>
#include<math.h>
#include<stdio.h>
//#include<cstdlib>

using namespace std;
const int maxn = 1100;

void mul(char a[],char b[],char c[])//c = a * b
{
    int  i,j,k,lena = strlen(a),lenb = strlen(b);
    int aa[maxn] = {0},bb[maxn] = {0},cc[maxn] = {0};

    int cnt,tt,temp;

    for( i = 0 ; i < lena; i++)
        aa[i] = a[lena -1 - i] - '0';
    for( i = 0 ; i < lenb; i++)
        bb[i] = b[lenb -1 - i] - '0';

    for(i = 0; i <lenb; i++)
    {
        cnt = 0;
        for(j = 0; j < lena; j++)
        {
            temp =  bb[i] * aa[j];
            tt= cc[i+j] + temp + cnt;
            cc[j+i] = tt % 10;
            cnt = tt / 10;
        }
        while(cnt != 0)
        {
            cc[j+i] = cnt % 10;
            cnt = cnt / 10;
            j++;
        }
    }

    for( k = 0; k < i + j - 1; k++)
        {c[k] = cc[i + j - k - 2] + '0';}
    c[k] = '';
}


void mi(char a[],int n,char b[]) // b = a ^ n
{
    int i;
    char c[maxn] = {''};
    strcpy(c,a);
    for( i = 1; i < n; i++)
    {
        mul(a,c,b);
        strcpy(c,b);
    }
}

int cmp(char a[],char b[]) // cmp a,b
{
    int lena = strlen(a),lenb = strlen(b);
    if(lena > lenb)
        return 1;
    else if(lena < lenb)
        return -1;
    else
        return strcmp(a,b);
}

void i2a(int x,char a[]) // int to array
{
    int i = 0,t,len,temp;
    while(x != 0)
    {
        a[i++] = x % 10 + '0';
        x = x / 10;
    }
    a[i] = '';
    len = strlen(a);
    for(i = 0 ; i < len / 2; i++)
    {
        temp = a[i];
        a[i] = a[len - i - 1];
        a[len - i -1] = temp;
    }
}
int main()
{
    //freopen("test.txt","r",stdin);
    //freopen("test1.txt","w",stdout);
    int n;
    double x0 = 1,x1 = pow(10,9),x2;

    char p[maxn] = {''},ans[maxn] = {''};
    char tt[maxn] = {''};

    int flag;

    while(scanf("%d%s",&n,p),(n != 0 && p[0] != 0))
    {
        if(n == 1)
        {
            printf("%sn",p);
            continue;
        }
        if(strcmp(p,"1")==0)
        {
            printf("%sn",p);
            continue;
        }
        x0 = 1,9);
        memset(ans,'',sizeof(ans));
        memset(tt,sizeof(tt));

        i2a(n,tt);

        while(1)
        {
            x2 = long (( x0 + x1 ) / 2);
            i2a(x2,tt);
<span style="white-space:pre">	</span>    mi(tt,n,ans);
           // cout<<"ans = "<<ans<<endl;
            flag = cmp(ans,p);
            if(flag < 0)
                x0 = x2;
            else if(flag > 0)
                x1 = x2 + 1;
            else if(flag == 0)
            {
                printf("%sn",tt);
                break;
            }
        }
    }
    return 0;
}

#include<stdio.h>
#include<math.h>
int main()
{
    double n,p,x;
    while(scanf("%lf%lf",&p)!=EOF)
    {
        if(n==0&&p==0) return 0;
        printf("%.lfn",pow(p,1.0/n));
    }
    return 0;
}

网上代码Java版

import java.util.Scanner;

public class Main{
	public static void main(String[] args){
		int n;
		double p;
		Scanner cin = new Scanner(System.in);
		while(true){
			n = cin.nextInt();
			p = cin.nextDouble();
			if(n == 0 && p == 0) break;
			System.out.printf("%.0fn",Math.pow(p,1.0/n));
		}
	}
}