c - 在C中将整数四舍五入到最接近的十或百

Question

我试图在 C 中考虑一个满足以下条件的函数：

它接受大于 0 的整数作为参数；
它将整数向上舍入到最接近的值，以便只有第一个数字不是零

例如：

53出来是60..

197出来是200..

4937 出来是 5000..

有没有办法做到这一点，无论尾随零的数量如何都能满足要求？

例如，我了解如何在任何个别情况下做到这一点。divide 53 by 10 then ceil(), multiply by 10,但我想要一个可以处理任何价值的人。

意见？想法？

score 20 · Accepted Answer

避免字符串转换和循环：

int num = ... // your number
int len = log10(num);
float div = pow(10, len);
int rounded = ceil(num / div) * div;

score 10 · Accepted Answer

It's unnecessary to convert the number to a string and back. You can do this using basic modulo arithmetic and multiplication and division.

Here's a pure numeric solution, hopefully somewhat more efficient in terms of running time:

int round_up_to_max_pow10(int n)
{
    int tmp = n;
    int i = 0;
    while ((tmp /= 10) >= 10) {
        i++;
    }

    if (n % (int)(pow(10, i + 1) + 0.5)) {
        tmp++;
    }

    for (; i >= 0; i--) {
        tmp *= 10;
    }

    return tmp;
}

printf("Original: %d; rounded: %d\n", 4937, round_up_to_max_pow10(4937));

score 3 · Accepted Answer

对数在这里非常有用，可以为“这有多少个零？”提供恒定时间的答案。

floor(log10(x))= z //the number of zeros

将取对数以 10 为底，并给出 x 中零的数量。

然后你可以使用 C 偶尔的成语

(A+B-1)/B

快速找到 A/B 的上限，从而以这种方式得出正确的前导数字：

zeros = exp10(1,z);
((x+zeros-1)/zeros) * zeros

这是伪代码，但你应该明白。关键的理解是对数是数学上确定一个数字有多少位数的方法。

score 1 · Accepted Answer

我不确定你想要圆形还是天花板。但是您在问题中表现出的行为暗示了 ceil。所以我把它包括在内。

int my_ceil(int num)
{
    int den = 1;
    int inc = 0;

    while (num >= 10) {
        inc += num % 10;
        num /= 10;
        den *= 10;
    }

    return (num + (inc > 0)) * den;
}

编辑

更改了代码以删除 ceil 和其他额外操作。

编辑 2

固定为 10 的倍数。

score 1 · Accepted Answer

通过 Cocoa API：

int number=9435;
NSString *string=[NSString stringWithFormat:@"%d",number];
long length=[string length];    
NSString *roundedString=[NSString stringWithFormat:@"%d",([[string substringToIndex:1]intValue]+1)];
while (--length>0) {
    roundedString=[roundedString stringByAppendingString:@"0"];
}
int roundedNumber=[roundedString intValue];
NSLog(@"%d,   %d",number,roundedNumber);

通过典型的 C 风格，数学上：

int i=8517;

int temp=i;
int len,msb;

for (len=0; temp>0; len++) {
    msb=temp;
    temp/=10;
}
msb++;
int multiplier=1;
for (int i=1; i<len; i++) {
    multiplier*=10;
}
 NSLog(@"Rounded : %d",msb*multiplier);

score 0 · Accepted Answer

这应该这样做：

static int
rnd_up(int val)
{
    double e, r;

    e = exp10(trunc(log10((double)val)));
    r = round(((double)val / e) + 0.5);

    return(r * e);
}

score 0 · Accepted Answer

Try taking the first character of the input number, add 1, then append zeros.

    Dim Input = "23568"
    Dim roundUp = Left(Input, 1) + 1

    For x = 1 To Len(Input) - 1
        roundUp &= "0"
    Next

In VB, but hopefully you get the idea.

score 0 · Accepted Answer

我会将数字转换为字符串。获取字符串的长度。

然后：

// pseudo code:
divisor = pow(10, len of number string)
answer = ceil (number/divisor) * divisor;

score 0 · Accepted Answer

没有循环。

#include <math.h>
unsigned roundToNextExp10( unsigned a )
{
    int d = a ;
    if( a >= 10 )
    {
        int m ;
        d-- ;
        m = (int)pow( 10, ((int)log10(d)) ) ;
        d = (int)((d / m) + 1) * m ;
    }        
    return d ;
}

score 0 · Accepted Answer

您可以将数字除以十，直到只剩下一位数字，然后将其乘回大小：

int n = 4937;

int m = 1;
while (n >= 10) {
  n = (n + 9) / 10;
  m *= 10;
}
n *= m;

score 0 · Accepted Answer

/*-------------------------------------------------*/
/* Round up a number                               */
/* (Negative numbers go DOWN)                      */
/* 34 -> 40                                        */
/* -34 -> -40                                      */
/*-------------------------------------------------*/
int round_up_10(int num) 
{
    int sign = 1;
    int tens = 1;

    if (num < 0)
    {
        num = (-1) * num;
        sign = -1;
    }
    num = (num + 9) / 10;
    num = num * 10;

    return sign * num;
}
/*-------------------------------------------------*/
/* Round down a number                             */
/* (Negative numbers go UP)                        */
/* 34 -> 30                                        */
/* -34 -> -30                                      */
/*-------------------------------------------------*/
int round_down_10(int num)   
{
    int sign = 1;
    int tens = 1;

    if (num < 0)
    {
        num = (-1) * num;
        sign = -1;
    }
    num = num / 10;
    num = num * 10;

    return sign * num;
}

main()
{
    printf("round_down_10(25450)= %d\n", round_down_10(25450));
    printf("round_down_10(-25450)= %d\n", round_down_10(-25450));
    printf("round_up_10(25450)= %d\n", round_up_10(25450));
    printf("round_up_10(-25450)= %d\n", round_up_10(-25450));

    printf("round_down_10(1347)= %d\n", round_down_10(1347));
    printf("round_down_10(-1347)= %d\n", round_down_10(-1347));
    printf("round_up_10(1347)= %d\n", round_up_10(1347));
    printf("round_up_10(-1347)= %d\n", round_up_10(-1347));

    printf("round_down_10(34)= %d\n", round_down_10(34));
    printf("round_down_10(-34)= %d\n", round_down_10(-34));
    printf("round_up_10(34)= %d\n", round_up_10(34));
    printf("round_up_10(-34)= %d\n", round_up_10(-34));
}
    /*
    The outout:
    round_down_10(25450)= 25450
    round_down_10(-25450)= -25450
    round_up_10(25450)= 25450
    round_up_10(-25450)= -25450
    round_down_10(1347)= 1340
    round_down_10(-1347)= -1340
    round_up_10(1347)= 1350
    round_up_10(-1347)= -1350
    round_down_10(34)= 30
    round_down_10(-34)= -30
    round_up_10(34)= 40
    round_up_10(-34)= -40
    */

score -1 · Accepted Answer

你在这里得到的一些专家建议......将一个数字四舍五入到最接近的 10：

if (i%10)
   i += 10 - i%10;

同样对于 100s、1000s 等。

c - 在C中将整数四舍五入到最接近的十或百

12 回答 12

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