题目

编写程序，读入几行文本，并打印一个表格，显示不同单词在文本中出现的次数。

考点

1. string到c中字符指针的转换
2. 使用strtok()拆解句子

实现

代码

#include<iostream>
#include<string>
#include<iomanip>
#pragma warning(disable:4996)
using namespace std;

const int MaxSize = 100;

//string到char指针需要手动转换，使用动态

int main() {
	//string word[MaxSize];
	string input = "to be or not to be , that is a question";
	//cin >> input;			//cin支持string嗷

	//string到char的动态转换
	int length = input.length();
	char *str = new char[length + 1];
	input.copy(str, length, 0);
	//第一个参数是接受的指针，第二个参数是复制的长度，第三个参数是开始位置可以不加，默认是从指针头
	str[length] = 0;

	//拆解单词
	char *p;
	char *word[MaxSize] = { 0, };
	p = strtok(str, " ");
	for (int i = 0; p != NULL; i++) {
		word[i] = p;
		p = strtok(NULL, " ");
	}

	string res[MaxSize];											//统计的单词列表
	int num[MaxSize] = { 0, };

	for (int i = 0, j = 0; word[i] != NULL; i++) {				//逐个对word中的单词进行处理
		bool a = (*word[i] >= ‘a‘ && *word[i] <= ‘z‘) || (*word[i] >= ‘A‘ && *word[i] <= ‘Z‘);
		if (a != 1)
			continue;
		int flag = 0;
		for (int k = 0; k < j; k++) {												//检查是否已经有
			if (res[k].compare(word[i]) == 0) {
				num[k]++;
				flag = 1;
			}
		}
		if (flag == 0) {
			res[j] = word[i];
			num[j]++;
			j++;
		}
	}


	for (int i = 0; num[i] != 0; i++)
		cout << res[i] << " " << setiosflags(ios_base::right)<<num[i] << endl;

	return 0;
}

输入

to be or not to be , that is a question

输出

to 2
be 2
or 1
not 1
that 1
is 1
a 1
question 1

读入几行文本并统计不同单词在文本中出现的次数

原文：https://www.cnblogs.com/Za-Ya-Hoo/p/12680683.html