算法实现c语言--02

1. 从键盘上输入字符，将小写字母转换成大写字母。输入“ctl + z” 结束 。

#include<stdio.h>
#include<stdlib.h>
//从键盘上输入字符，将小写字母转换成大写字母。输入“ctl + z” 结束 

int main()
{
    char c;
    while (fflush(stdin),scanf_s("%c", &c) != EOF)
    {
        if (c >= 'a'&&c<='z' )
        {
            c = c - 32;
            printf("%c", c);
            continue;
        }
        if (c >= 'A'&&c < ='Z')
        {
            printf("%c", c);
            continue;
        }
        else printf("error,input agian
");
    }

    printf("
");
    system("pause");
    return 0;
}

1. 从键盘上输入字符，（1）分别统计一下其中字母，数字，其他字符的个数， （2）将统计的字母，数字，其他字符的个数以柱状图的形式打印。例如 

  5

*****

*****     3

*****   *****     2

*****   *****   *****

*****   *****   *****

 alp     num     oth

 

#include<stdio.h>
#include<stdlib.h>
typedef struct {
    int n;
    char *name;
}elem;
elem alp, num, oth, print[3], tmp;
void paixu(elem *print)
{
    if ((print[0].n < print[1].n) == 1)
    {
        tmp = print[0];
        print[0] = print[1];
        print[1] = tmp;
    }
    if ((print[0].n < print[2].n) == 1)
    {
        tmp = print[0];
        print[0] = print[2];
        print[2] = tmp;
    }
    if ((print[1].n < print[2].n) == 1)
    {
        tmp = print[1];
        print[1] = print[2];
        print[2] = tmp;
    }
}
int main()
{
    char c;
    int cnt[3] = { 0 };
    int i = 0, j = 0;
    int maxc;
    alp.n = 0;
    num.n = 0;
    oth.n = 0;
    alp.name = "alp";
    num.name = "num";
    oth.name = "oth";

    int flag = 1;
    while (fflush(stdin), scanf_s("%c", &c) != EOF)
    {
        if (c >= 'a'&&c <= 'z' || c >= 'A'&&c <= 'Z')
        {
            alp.n++;
            printf("%c", c);
            continue;
        }
        else if (c >= 48 && c <= 57)
        {
            num.n++;
            printf("%c", c);
            continue;
        }
        else if (c != '') {
            oth.n++;
            printf("%c", c);
        }
        if (c == '
')break;
    }
    oth.n--;


    print[0] = alp;
    print[1] = num;
    print[2] = oth;
    paixu(print);

    for (i = 0; i <= print[0].n; i++)
    {
        printf("
");
        if (print[0].n - i <= print[0].n)
        {

            if (print[0].n - i == print[0].n)
            {
                if (print[0].n == print[1].n)
                {
                    if (print[1].n == print[2].n)
                        printf("%s	%s	%s", print[0].name, print[1].name, print[2].name);
                    else printf("%s	%s", print[0].name, print[1].name);
                }
                else printf("%s", print[0].name);
                continue;
            }
            printf("*****	");
            //////////////////////////////////
            if (print[0].n - i <= print[1].n)
            {
                if (print[0].n - i == print[1].n)
                {
                    if (print[1].n == print[2].n)
                        printf("%s	%s", print[1].name, print[2].name);
                    else printf("%s", print[1].name);
                    continue;
                }
                printf("*****	");
                /////////////////////////////////////
                if (print[0].n - i <= print[2].n)
                {
                    if (print[0].n - i == print[2].n)
                    {
                        printf("%s", print[2].name);
                        continue;
                    }
                    printf("*****	");
                }
            }
        }
    }
    printf("
");

    printf("%d	", print[0].n);
    printf("%d	", print[1].n);
    printf("%d	", print[2].n);
    printf("
");


    printf("
");
    system("pause");
    return 0;
}

3.进制转换。

（1） 将十进制数转换成二进制数。输入十进制数输出对应的二进制数，输入“ctl + z”结束。

 

#include<stdio.h>
#include<stdlib.h>

#define N 8
void fun(int num, int k)
{
    int arr[N], i;
    for (i = 0; i <N; i++)
    {
        arr[i] = num % k;
        num = num / k;
        if (num == 0)
            break;
    }
    for (; i >= 0; i--)
        printf("%d", arr[i]);
    printf("


");
}

int main()
{
    int n;

    while (fflush(stdin), scanf_s("%d", &n) != EOF)
    {
        fun(n, 2);
    }
    printf("
");
    system("pause");
    return 0;
}

 

(2)将二进制转换成十进制数。输入二进制数输出对应的十进制数，输入“ctl + z”结束

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

int main()
{
    char *p, s[6]; int n;
    p = s;
    gets(p);
    n = 0;
    while (*p != '')
    {
        n = n * 2 + *p - '0';
        p++;
    }
    printf("%d", n);
    system("pause");
    return 0;
}

（3） 将十进制数转换成十六进制数。输入十进制数输出对应的十六进制数。输入“ctl + z” 结束。

 

#include<stdio.h>
#include<stdlib.h>

#define N 8
void fun(int num)
{
    int arr[N], i;
    for (i = 0; i <N; i++)
    {
        arr[i] = num % 16;
        num = num / 16;
        if (num == 0)
            break;
    }
    for (; i >= 0; i--)
        switch (arr[i])
        {
        case 10: printf("A"); break;
        case 11: printf("B"); break;
        case 12: printf("C"); break;
        case 13: printf("D"); break;
        case 14: printf("E"); break;
        case 15: printf("F"); break;
        default: printf("%d", arr[i]);
        }
    printf("


");
}

int main()
{
    int n;

    while (fflush(stdin), scanf_s("%d", &n) != EOF)
    {
        fun(n);
    }
    printf("
");
    system("pause");
    return 0;
}

 

（4） 将十六进制数转换成十进制数。输入十六进制数输出对应的十进制数。输入“ctl +  z” 结束。

 

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

int main()
{
    char *p, s[6]; int n;
    p = s;
    gets(p);
    n = 0;
    
    while (*p != '')
    {
        switch (*p)
        {
        case '0':n = 0; break;
        case '1':n = 1; break;
        case '2':n = 2; break;
        case '3':n = 3; break;
        case '4':n = 4; break;
        case '5':n = 5; break;
        case '6':n = 6; break;
        case '7':n = 7; break;
        case '8':n = 8; break;
        case '9':n = 9; break;
        case 'A':n = 10; break;
        case 'B':n = 11; break;
        case 'C':n = 12; break;
        case 'D':n = 13; break;
        case 'E':n = 14; break;
        case 'F':n = 15; break;
        default:break;

        }
        n = n * 16 + *p - '0';
        p++;
    }
    printf("%d", n);
    system("pause");
    return 0;
}

 

4.统计一个整数对应的二进制数的1的个数。输入一个整数（可正可负）， 输出该整数的二进制包含1的个数， “ctl+ z” 结束。

#include<stdio.h>
#include<stdlib.h>

#define N 8
void fun(int num, int k)
{
    int arr[N], i, cnt = 0;
    for (i = 0; i <N; i++)
    {
        arr[i] = num % k;
        num = num / k;
        if (num == 0)
            break;
    }
    for (; i >= 0; i--)
    {
        if (arr[i] == 1)cnt++;
        printf("%d", arr[i]);
    }
    printf("共有%d个1",cnt);
    printf("


");
}

int main()
{
    int n;

    while (fflush(stdin), scanf_s("%d", &n) != EOF)
    {
        fun(n, 2);
    }
    printf("
");
    system("pause");
    return 0;
}

5.有101个整数，其中有50个数出现了两次，1个数出现了一次， 找出出现了一次的那个数。

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

int main()
{
    int a[101];
    int i, ret = 0;
    for (i = 0; i < 50; i++) a[i] = i + 1;
    for (i = 51; i <= 100; i++) a[i] = i - 50;
    scanf_s("%d", &a[50]);
    for (i = 0; i <= 100; ++i)
    {
        ret = ret ^ a[i];
    }
    printf("%d
", ret);
    printf("
");
    system("pause");
    return 0;
}

6.

（1）输入年月日，输出该日期是当年的第几天。

#include<stdio.h>
#include<stdlib.h>

int main()
{
    int a[12] = {31,28,31,30,31,30,31,31,30,31,30,31 };
    int year, mon, day,sum=0;
    scanf_s("%d%d%d", &year, &mon, &day);
    for (int i = 0; i < mon - 1; i++)sum += a[i];
    sum += day;
    if (mon > 2)sum =sum+(year % 4 && year % 100 || year % 400);
    printf("%d", sum);

    printf("
");
    system("pause");
    return 0;
};

（2）输入两个日期（年 月 日 年 月 日）， 输出这两个日期之间差多少天

#include<stdio.h>
#include<stdlib.h>

int isleap(int year)
{
    return year % 4==0 && year % 100==0 || year % 400==0;
}
int main()
{
    int a[12] = { 31,28,31,30,31,30,31,31,30,31,30,31 };
    int year1 = 0, mon1 = 0, day1 = 0, year2 = 0, mon2 = 0, day2 = 0, sum1 = 0, sum2 = 0, sum = 0, gap = 0;

    scanf_s("%d %d %d", &year1, &mon1, &day1);
    scanf_s("%d %d %d", &year2, &mon2, &day2);

    for (int i = 0; i < mon1 - 1; i++)sum1 += a[i];
    sum1 += day1;
    if (mon1 > 2)sum1 = sum1 + isleap(year1);
    printf("%d天", sum1);

    for (int i = 0; i < mon2 - 1; i++)sum2 += a[i];
    sum2 += day2;
    if (mon2 > 2)sum2 = sum2 + isleap(year2);
    printf("%d天", sum2);

    gap = abs(year2 - year1);
    sum = sum2 - sum1 + gap * 365 + gap / 4 ;

    printf("%d天", sum);

    printf("
");
    system("pause");
    return 0;
}

（3）输入一个日期，输出该日期是星期几。

#include<stdio.h>
#include<stdlib.h>

int isleap(int year)
{
    return year % 4==0 && year % 100==0 || year % 400==0;
}
int main()
{
    int a[12] = { 31,28,31,30,31,30,31,31,30,31,30,31 };
    int year1 = 2018, mon1 = 3, day1 = 4, year2 =0 , mon2 = 0, day2 = 0, sum1 = 0, sum2 = 0, sum = 0, gap = 0;
    int week = 0;

    scanf_s("%d %d %d", &year2, &mon2, &day2);

    for (int i = 0; i < mon1 - 1; i++)sum1 += a[i];
    sum1 += day1;
    if (mon1 > 2)sum1 = sum1 + isleap(year1);
    for (int i = 0; i < mon2 - 1; i++)sum2 += a[i];
    sum2 += day2;

    if (mon2 > 2)sum2 = sum2 + isleap(year2);

    gap = abs(year1 - year2);
    sum = sum2 + gap * 365 + gap / 4 - sum1;
    week = sum;
    while (week /7!=0)
    {
        week %= 7;
    }
    if(week==0||week==7)
        printf("星期日", week);
    else
        printf("星期%d", week);

    printf("
");
    system("pause");
    return 0;
}

4.输入 一个日期 和一个整数 n，输出从该日期起经过n天以后的日期。

 -> date -> tomorrow of the date

#include<stdio.h>
#include<stdlib.h>

int isleap(int year)
{
    return year % 4 == 0 && year % 100 == 0 || year % 400 == 0;
}
int main()
{
    int a[12] = { 31,28,31,30,31,30,31,31,30,31,30,31 };
    int year1 = 2018, mon1 = 3, day1 = 5, year2 = 0, mon2 = 0, day2 = 0, sum1 = 0, sum2 = 0, sum = 0,
        gap = 1, days = 0;
    int week = 0;
    int i;
    scanf_s("%d", &days);

    for (i = 0; i < mon1 - 1; i++)sum1 += a[i];
    sum1 += day1;
    if (mon1 > 2)sum1 = sum1 + isleap(year1);

    sum = sum1 + days;
    year2 = year1;

    while (sum > 365 + isleap(year2))
    {
        year2 = year2 + gap;
        sum -= 365 - isleap(year2);
        
    } ;

    if (sum < 365 + isleap(year2))
    {
        for (i = 0; sum > a[i]; i++)
        {
            sum -= a[i];
        }
        i++;
        mon2 = i;
        day2 = sum;
        printf("%d年%d月%d日", year2, mon2, day2);
    }

        printf("
");
    system("pause");
    return 0;
}