C/C++题库整理（二）

题目一：

/*写出一个程序，接受一个正浮点数值，输出该数值的近似整数值。如果小数点后数值大于等于5,向上取整；小于5，则向下取整。#include <stdio.h>#include <string.h>#include <stdlib.h>int main(){char float_num[200];int flag_dian = 0;int flag_jinwei = 0;int ans = 0;while (scanf("%s", &float_num) != EOF){for (int i = 0; i < strlen(float_num); i++){if (float_num[i] == '.') flag_dian = i;}if (float_num[flag_dian + 1] >= ('5')) {flag_jinwei = 1;}else flag_jinwei = 0;float_num[flag_dian] = '\0';ans = atoi(float_num);printf("%d", ans + flag_jinwei);}return 0;}*/

题目二：

/*输入一个int型整数，按照从右向左的阅读顺序，返回一个不含重复数字的新的整数。输入描述:输入一个int型整数输出描述:按照从右向左的阅读顺序，返回一个不含重复数字的新的整数示例1输入9876673输出37689#include <stdio.h>int main(void) {int num = 0,i = 0;int flag[10] = {0,0,0,0,0,0,0,0,0,0};//用flag[3]来做3这个字符的flag，妙！while (scanf("%d",&num) != EOF){while(num){int current = num % 10;if (flag[current] != 1){flag[current] = 1;printf("%d", current);}num = num / 10;}}return 0;}*/

题目三：

/*编写一个函数，计算字符串中含有的不同字符的个数。字符在ACSII码范围内(0~127)。不在范围内的不作统计。*/#include <stdio.h>#include <string.h>int main(void) {char str[1000];while (scanf("%s",&str) != EOF){int flag[128] = { 0 };int length = strlen(str);int count = 0;int k = 0;int i = 0;while (i<length){k = str[i];if ((flag[k] == 0)&&(k<=127)&&(k>=0)){count++;flag[k] = 1;}i++;}printf("%d\n", count);}return 0;}