C++博客-purplest-随笔分类-高精度

poj 1001 Exponentiation

purplest — Tue, 04 Oct 2011 11:55:00 GMT

求实数的冥，虽然题没啥难度，为了弄下java做的

java就是方便啊，几行就搞定了c++近100行的东东，哎，为嘛STL不封装biginteger呢？

import java.math.BigDecimal;
import java.util.Scanner;

public class Main {
    public void slove()
    {
        Scanner cin = new Scanner(System.in);
        while ( cin.hasNext() )
        {
            BigDecimal num = cin.nextBigDecimal();
            int n = cin.nextInt();
            BigDecimal sum = new BigDecimal("1");
            String str;
            while ( n > 0 )
            {
                if ( ( n & 1 ) > 0 )
                    sum = sum.multiply(num);
                num = num.multiply(num);
                n>>=1;
            }
            str= sum.stripTrailingZeros().toPlainString();
            if ( str.startsWith("0.") )
                str= str.substring(1);
            System.out.println(str);
        }
    }
    public static void main(String[] args) {
        new Main().slove();
    }
}

purplest 2011-10-04 19:55 发表评论

poj 1220 NUMBER BASE CONVERSION

purplest — Fri, 12 Aug 2011 01:48:00 GMT

高精度的进制转换

#include<stdio.h>
#include<string.h>

int n, ori, to, t[1000], tonum[1000];
char str[1000], num[1000];

int main ()
{
    scanf("%d", &n);
    while (n--)
    {
        scanf("%d%d%s", &ori, &to, str);
        int i, j= 0, len= strlen(str), k= len;
        for (i=len-1; 0 <= i; i--)
            t[len-1-i]= str[i]-( 58 > str[i] ? '0' : 91 > str[i] ? 'A'-10 : 'a'-36 );
        while( k )
        {
            for ( i = k-1 ; i > 0; i-- )
            {
                t[i-1]+=t[i]%to*ori;
                t[i]/=to;
            }
            tonum[j++]=t[0]%to;
            t[0]/=to;
            while( 0 < k && !t[k-1]) k--;
        }
        for ( num[j] = i = 0 ; i < j ; i++ )
            num[j-1-i]=tonum[i]+( 10 > tonum[i] ? '0' : 36 > tonum[i] ? 'A'-10 :'a'-36);
        printf("%d %s\n%d %s\n\n", ori, str, to, num);
    }
    return 0;
}

purplest 2011-08-12 09:48 发表评论

poj 2845 01000001

purplest — Wed, 10 Aug 2011 16:45:00 GMT

高精度模拟

#include <stdio.h>
#include <string.h>

int n;
char num1[100], num2[100],num[100];

void turn(char *str, int len)
{
    int mid=len>>1;
    char temp;
    for ( int i= 0; i < mid; i++ )
    {
        temp= str[i];
        str[i]=str[len-1-i];
        str[len-1-i]= temp;
    }
}

int calu(int len1, int len2)
{
    int max, min;
    char *str;
    if (len1 > len2)
    {
        max=len1;
        min=len2;
        str=num1;
    }
    else
    {
        max= len2;
        min= len1;
        str=num2;
    }
    int i;
    memset(num, 0, sizeof(num));
    for ( i = 0 ; i < min; i++ )
    {
        num[i]+=num1[i]+num2[i];
        if (num[i]>1)
        {
            num[i+1]= num[i]/2;
            num[i]%=2;
        }
    }
    for ( i = min; i < max; i++ )
    {
        num[i]+=str[i];
        if (num[i]>1)
        {
            num[i+1]= num[i]/2;
            num[i]%=2;
        }
    }
    if (num[i]) return i;
    else
        for ( i=i-1; i >0; i-- )
            if (num[i]) return i;
    return 0;
}

int main()
{
    scanf("%d", &n);
    int l;
    for ( l = 1 ; l <= n ; l++ )
    {
        scanf("%s%s", num1, num2);
        printf("%d ", l);
        int len1= strlen(num1), len2= strlen(num2);
        int i;
        for ( i = 0 ; i < len1; i++ )
            num1[i]-='0';
        for ( i = 0 ; i < len2; i++ )
            num2[i]-='0';
        turn(num1, len1);
        turn(num2, len2);
        int len= calu(len1, len2);
        for ( i = len ; i >= 0; i-- )
            printf("%d", num[i]);
        putchar(10);
    }
    return 0;
}

purplest 2011-08-11 00:45 发表评论

poj 1047 Round and Round We Go

purplest — Fri, 05 Aug 2011 15:21:00 GMT

枚举下，再验证，高精度乘法那需要处理好，0可以在最前端

#include <stdio.h>
#include <string.h>

char str[100];
char num[100];

void turn(int len)
{
    int mid=len>>1;
    char temp;
    for ( int i= 0; i < mid; i++ )
    {
        temp= str[i];
        str[i]=str[len-1-i];
        str[len-1-i]= temp;
    }
}

int calu(int k, int len)
{
    memset(num, 0, sizeof(num));
    int i, t;
    for ( i = 0; i < len; i++ )
    {
        t= k*str[i];
        if (t <= 9)
        {
            num[i]+=t;
            if (num[i]>9)
            {
                num[i+1]=num[i]/10;
                num[i]%=10;
            }
        }
        else
        {
            num[i]+=t%10;
            if (num[i]>9)
            {
                num[i+1]=num[i]/10;
                num[i]%=10;
            }
            num[i+1]+=t/10;
        }
    }
    if (num[i+1]) i++;
    return i;
}

int cmp(int len1, int len2)
{
    int i, j;
    if (len2 != len1 ) return 0;
    for ( i = 0; i < len1; i++)
    {
        if (str[i]==num[0])
        {
            for ( j = 0; (j < len1 ) ; j++ )
                if (str[(i+j)%len1] != num[j%len2])
                    break;
            if ( j == len1 ) return 1;
        }
    }
    return 0;
}

int main()
{
    while (EOF != scanf("%s", str))
    {
        printf("%s is ", str);
        int len= strlen(str), i, flag=1;
        for ( i = 0; i < len; i++ )
        {
            str[i]-='0';
        }
        turn(len);
        for ( i = 1; i <= len; i++ )
        {
            int len2= calu(i, len);
            if ( !cmp(len, len2) )
            {
                printf("not cyclic\n");
                flag=0;
                break;
            }
        }
        if ( flag ) printf("cyclic\n");
    }
    return 0;
}

purplest 2011-08-05 23:21 发表评论