2016年4月1日C语言作业答案及遇到的问题

题目及解答

2.一年中的第几天——改编自习题4.3

原题链接

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#include<stdio.h>;
int main (void)  
{
    int year, month, day;
    int count;
    int isLeap;
    while (scanf("%d%d%d", &year, &month, &day) == 3)
    {
        count = 0;
        isLeap = (year % 4 == 0 && year % 100 != 0) || (year % 100 == 0 && year % 400 == 0);
        count += day;
        month -= 1;
        switch (month)
        {
            case 12: count += 31; month -= 1;
            case 11: count += 30; month -= 1;
            case 10: count += 31; month -= 1;
            case 9: count += 30; month -= 1;
            case 8: count += 31; month -= 1;
            case 7: count += 31; month -= 1;
            case 6: count += 30; month -= 1;
            case 5: count += 31; month -= 1;
            case 4: count += 30; month -= 1;
            case 3: count += 31; month -= 1;
            case 2: count += 28 + isLeap; month -= 1;
            case 1: count += 31; month -= 1;
            default: break;
        }
        printf("该日期是这一年中的第%d天\n", count);
    }
    return 0;
}

3.税金额度计算——改编自习题4.4

原题链接

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/* ZhihaoChen CSEE 1501 */
#include<stdio.h>;
/* Prototype Declaration */
void if_else_process (double x);
void switch_process (double x);
/* Main Function */
int main(void)
{
    int number;
    scanf("%d", &number);
    int t; double x;
    int count;
    for (count = 0 ; count & number ; ++count)
    {
        scanf("%d%lf", &t, &x);
        switch (t)
        {
            case 0:
                if_else_process(x);
                break;
            case 1:
                switch_process(x);
                break;
            case 2:
                if_else_process(x);
                switch_process(x);
                break;
            default:
                break;
        }
        if (count != (number - 1))
            printf("\n");
    }
    return 0;
}
/* data processing and output */
void if_else_process (double x)
{
    double output;
    if (x < 1000)
        output = 0;
    else if (x >= 1000 && x < 2000)
        output = x * 0.05;
    else if (x >= 2000 && x < 3000)
        output = x * 0.10;
    else if (x >= 3000 && x < 4000)
        output = x * 0.15;
    else if (x >= 4000 && x < 5000)
        output = x * 0.20;
    else if (x > 5000)
        output = x * 0.25;
    printf ("After if-else processing,the amount of tax to be paid is : %.2f\n", output);
}

void switch_process (double x)
{
    double output;
    switch((int)x / 1000)
    {
        case 0: output = 0; break;
        case 1: output = x * 0.05; break;
        case 2: output = x * 0.10; break;
        case 3: output = x * 0.15; break;
        case 4: output = x * 0.20; break;
        default: output = x * 0.25; break;
    }
    printf("After switch processing,the amount of tax to be paid is : %.2f\n", output);
}

4.两个实数的四则运算——改编自习题4.5

原题链接

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/* ZhihaoChen CSEE 1501 */
#include<stdio.h>;
/* Prototype Declaration */
void if_else_process(double a, double b, char c);
void switch_process(double a, double b, char c);
/* Main Function */
int main(void)
{
    int t, count;
    double a, b;
    char c;
    for (count = 0 ; (scanf("%d%lf%lf %c", &t, &a, &b, &c) != EOF) ; ++count)
    {
        if(count != 0)
            printf("\n");
        switch (t)
        {
            case 0:
                if_else_process(a, b, c);
                break;
            case 1:
                switch_process(a, b, c);
                break;
            case 2:
                if_else_process(a, b, c);
                switch_process(a, b, c);
                break;
            default:
                break;
        }
    }
    return 0;
}
/* data processing and output */
void if_else_process (double a, double b, char c)
{
    double output = 0.0;
    if (c == '+')
        output = a + b;
    if (c == '-')
        output = a - b;
    if (c == '*')
        output = a * b;
    if (c == '/')
        output = a / b;
    printf ("After if-else processing,the result is : %.2f\n", output);
}

void switch_process (double a, double b, char c)
{
    double output = 0.0;
    switch(c)
    {
        case '+': output = a + b; break;
        case '-': output = a - b; break;
        case '*': output = a * b; break;
        case '/': output = a / b; break;
        default: break;
    }
    printf("After switch processing,the result is : %.2f\n", output);
}

5.合并空格字符输出——改编自习题4.7

原题链接

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#include <stdio.h>
#define CHAR 0
#define SPACE 1

int main (void)
{
    char character;
    int status = CHAR;
    while ((character = getchar()) != EOF)
    {   
        if ((character == ' ') && (status == CHAR))
        {
            status = SPACE;
            printf(" ");
        }
        if ((character != ' ') && (status == CHAR))
        {
            printf("%c", character);
        }
        if ((character != ' ') && (status == SPACE))
        {
            status = CHAR;
            printf("%c", character);
        }
    }
    return 0;
}

6.求π的近似值——改编自习题4.10

原题链接

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#include <stdio.h>
#include <math.h>

int main (void)
{
    double item = 1.0;
    double pi = 0.0;
    int count;
    for (count = 0 ; (item >= 1.0e-5) ; ++count)
    {
        item = 1.0/(2 * count + 1);
        if (count % 2 == 0)
            pi += item;
        else
            pi += (-item);
    }
    printf("%.9f\n", pi * 4);
    return 0;
}

7.验证哥德巴赫猜想——改编自习题4.13

原题链接

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#include <stdio.h>
#include <math.h>

int isPrime (int x);

int main (void)
{
    int number, testnum;
    scanf ("%d", &number);
    int count, subnumber1, subnumber2;
    for (count = 0 ; count < number ; ++count)
    {
        scanf("%d", &testnum);
        for (subnumber1 = 2 ; subnumber1 <= (testnum / 2) ; ++subnumber1)
        {
            subnumber2 = testnum - subnumber1;
            if (isPrime(subnumber2) && isPrime(subnumber1))
            {
                printf("%d=%d+%d\n", testnum, subnumber1, subnumber2);
                break;
            }
        }
    }
    return 0;
}

int isPrime (int x)
{
    int i;
    for (i = 2 ; i <= sqrt(x) ; ++i)
        if (x % i == 0) return 0;
    return 1;
}

8.正整数相加求平均数——改编自习题4.17

原题链接

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#include <stdio.h>
int main (void)
{
    int number;
    scanf("%d", &number);
    int count, subcount, status; int n, positiveNumber;
    long sumNumber;
    double averageNumber;
    for (count = 0 ; count < number ; ++count)
    {
        n = 0; sumNumber = 0;
        positiveNumber = 0;
        averageNumber = 0.0;
        scanf("%d", &status);
        switch(status)
        {
            case 0:
                for (subcount = 0 ; subcount < 10 ; ++subcount)
                {
                    scanf("%d", &n);
                    if (n <= 0) continue;
                    sumNumber += n;
                    positiveNumber += 1;
                }
                averageNumber = (double)sumNumber / positiveNumber;
                if (positiveNumber != 0)
                    printf("In \"continue\" way,numbers=%d,average=%f\n", positiveNumber, averageNumber);
                break;
            case 1:
                for (subcount = 0 ; subcount < 10 ; ++subcount)
                {
                    scanf("%d", &n);
                    if(n > 0)
                    {
                        sumNumber += n;
                        positiveNumber += 1;
                    }
                }
                averageNumber = (double)sumNumber / positiveNumber;
                if (positiveNumber != 0)
                    printf("In \"no continue\" way,numbers=%d,average=%f\n", positiveNumber, averageNumber);
                break;
            default:
            break;
        }
    }
    return 0;
}

要注意的问题

1.注意审题，比如第六题注意到“最后一项”的绝对值，再比如说第三题注意到输出格式逗号前后都没有空格

2.注意数据类型的范围

3.注意浮点数的科学计数法表示

4.注意条件运算要不要加等于号

5.逻辑错误比语法错误更难检查，要关注程序的细节

6.注意scanf在读取的字符(%c)时候会识别’ ‘，所以在格式化字符串中用’ %c’或者’%1s’均可（后者无法通过系统）

7.注意浮点数的运算存在精度损失，例：double i = 1/e-5 而(int)i＝99999,在作为循环条件时需要格外注意，一般情况下最好避免，或者+0.0001再取整