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
     
      #include
      
       //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\n");    }    printf("\n");    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
     
      #include
      
       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 != '\0') {            oth.n++;            printf("%c", c);        }        if (c == '\n')break;    }    oth.n--;    print[0] = alp;    print[1] = num;    print[2] = oth;    paixu(print);    for (i = 0; i <= print[0].n; i++)    {        printf("\n");        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\t%s\t%s", print[0].name, print[1].name, print[2].name);                    else printf("%s\t%s", print[0].name, print[1].name);                }                else printf("%s", print[0].name);                continue;            }            printf("*****\t");            //            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\t%s", print[1].name, print[2].name);                    else printf("%s", print[1].name);                    continue;                }                printf("*****\t");                /                if (print[0].n - i <= print[2].n)                {                    if (print[0].n - i == print[2].n)                    {                        printf("%s", print[2].name);                        continue;                    }                    printf("*****\t");                }            }        }    }    printf("\n");    printf("%d\t", print[0].n);    printf("%d\t", print[1].n);    printf("%d\t", print[2].n);    printf("\n");    printf("\n");    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
     
      #include
      
       #define N 8void fun(int num, int k){    int arr[N], i;    for (i = 0; i 
       
        = 0; i--)        printf("%d", arr[i]);    printf("\n\n\n");}int main(){    int n;    while (fflush(stdin), scanf_s("%d", &n) != EOF)    {        fun(n, 2);    }    printf("\n");    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
     
      #include
      
       #include
       
        int main(){    char *p, s[6]; int n;    p = s;    gets(p);    n = 0;    while (*p != '\0')    {        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
     
      #include
      
       #define N 8void fun(int num){    int arr[N], i;    for (i = 0; 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("\n\n\n");}int main(){    int n;    while (fflush(stdin), scanf_s("%d", &n) != EOF)    {        fun(n);    }    printf("\n");    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
     
      #include
      
       #include
       
        int main(){    char *p, s[6]; int n;    p = s;    gets(p);    n = 0;        while (*p != '\0')    {        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
     
      #include
      
       #define N 8void fun(int num, int k){    int arr[N], i, cnt = 0;    for (i = 0; i 
       
        = 0; i--)    {        if (arr[i] == 1)cnt++;        printf("%d", arr[i]);    }    printf("共有%d个1",cnt);    printf("\n\n\n");}int main(){    int n;    while (fflush(stdin), scanf_s("%d", &n) != EOF)    {        fun(n, 2);    }    printf("\n");    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
     
      #include
      
       #include
       
        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\n", ret);    printf("\n");    system("pause");    return 0;}
       
      
     

6.

（1）输入年月日，输出该日期是当年的第几天。

 

#include
     
      #include
      
       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("\n");    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
     
      #include
      
       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("\n");    system("pause");    return 0;}
      
     

 

（3）输入一个日期，输出该日期是星期几。

 

 

#include
     
      #include
      
       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("\n");    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
     
      #include
      
       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("\n");    system("pause");    return 0;}