Fractions to Decimals（模拟）

解析分数转换为小数并标记循环节的算法实现

最新推荐文章于 2023-12-03 11:51:16 发布

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本文详细介绍了如何通过编程解决分数转换为小数的问题，并特别关注了循环节的标记方法。文章包括输入输出格式、算法思路、代码实现以及输出样例，旨在提供一种有效的方法来解决数学中的分数转换问题。

Fractions to Decimals

Write a program that will accept a fraction of the form N/D, where N is the numerator and D is the denominator and print the decimal representation. If the decimal representation has a repeating sequence of digits, indicate the sequence by enclosing it in brackets. For example, 1/3 = .33333333...is denoted as 0.(3), and 41/333 = 0.123123123...is denoted as 0.(123). Use xxx.0 to denote an integer. Typical conversions are:

1/3     =  0.(3)
22/5    =  4.4
1/7     =  0.(142857)
2/2     =  1.0
3/8     =  0.375
45/56   =  0.803(571428)

PROGRAM NAME: fracdec

INPUT FORMAT

A single line with two space separated integers, N and D, 1 <= N,D <= 100000.

SAMPLE INPUT (file fracdec.in)

45 56

OUTPUT FORMAT

The decimal expansion, as detailed above. If the expansion exceeds 76 characters in length, print it on multiple lines with 76 characters per line.

SAMPLE OUTPUT (file fracdec.out)

0.803(571428)

题意：

给出 N，D（1 ~ 100000），代表分子和分母，求出商，循环节用括号括起来。每76个字符输出一行。

思路：

模拟。用 res 数组保留商，rem 数组标记余数的位置（数组标记位置法），若出现一样的余数说明有循环节，记录这个位置到时候输出“（" 用。

AC：

/*
TASK:fracdec
LANG:C++
ID:sum-g1
*/
#include <cstdio>
#include <string.h>
#define MAX 500000
using namespace std;

int res[MAX], rem[100005];
//rem记录位置，res记录结果
int main() {
    freopen("fracdec.in", "r", stdin);
    freopen("fracdec.out", "w", stdout);

    int den, num, r = -1;
    int ans = 0, same = -1, k = 0;

    scanf("%d%d",&den,&num);
    memset(rem,-1,sizeof(rem));

    while (r && same == -1) {
            r = den % num;
            rem[r] != -1 ? same = rem[r] : rem[r] = k++;
            res[ans++] = den / num;
            den = r * 10;
    }

    int time = 0, fir = res[0];
    printf("%d.", res[0]);

    if (!fir) ++time;
    while (fir) {
        ++time;
        fir /= 10;
    }
    ++time;

    if (same == -1) {
            for (int i = 1; i < ans; ++i) {
                    printf("%d", res[i]);
                    if(!(++time % 76)) printf("\n");
            }

            if(k == 1) {
                    printf("0");
                    if(!(++time % 76)) printf("\n");
            }
            printf("\n");
    }
    else {
            for (int i = 0; i < ans; ++i) {
                    if(i) {
                            printf("%d",res[i]);
                            if(!(++time % 76)) printf("\n");
                    }
                    if (i == same) {
                            printf("(");
                            if(!(++time % 76)) printf("\n");
                    }
            }
            printf(")\n");
    }
    return 0;
}