- 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\n");}printf("\n");system("pause");return 0; }
- 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 != '\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<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("\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<stdio.h> #include<stdlib.h> #include<string.h>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<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("\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<stdio.h> #include<stdlib.h> #include<string.h>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<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("\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<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\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<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("\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<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("\n");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);elseprintf("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<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("\n");system("pause");return 0; }