ACM练习题C++版答案记录-Section 2

最近复习C++，拿起ACM从头练，在此做个答案记录，大家不要盲目复制哦，要有自己的思考哦

练习地址：http://acm.hdu.edu.cn/userloginex.php

//1.2.1 Text Reverse 变量绕来绕去的，注意不要粗心就是了
#include<string>
#include<stdio.h>
#include<iostream>
using namespace std;
int main(){
	int k;
	cin >> k;
	getchar(); // because '\n' also is a string type, so need a getchar here 
	string s;
	while(k--){
		
		getline(cin, s);
		int start = 0, end;
		int len = s.length();
		for(int i=0; i <= len; i++){
			if(s[i] == ' ' || i == len){
				end = i-1;
				for(int j = end; j >= start; j--){
					cout<<s[j];
				}
				start = i + 1;
				if(i != len){
					cout<< " ";
				}else{
					cout<<endl;
				}
			}
		}
	}
	return 0;
}

// 1.2.2 Financial Management 这里用了一个floor方法
#include<Math.h>
#include<iostream>
using namespace std;

int main()
{
	int n = 12;
	double a;
	double sum = 0.0;
	int i = n;
    while(i--)
    {
    	cin>>a;
    	sum += a;
    }
    int res = 0;
    double mean = sum * 100 / n;
    int floor_mean = floor(mean);
    if(mean - floor_mean < 0.5){
    	res = floor_mean;
	}else if(mean - floor_mean >= 0.5){
		res = floor_mean + 1;
	}
	int front = res / 100;
	int behind = res - front * 100;
	cout<< "$" << front << "." << behind<<endl;
    return 0;
}

//1.2.3 An Easy Task 关于闰年的，有点巧妙
#include<iostream>
using namespace std;
int main()
{
	int n, year, after, k;
	cin>>n;
	while(n--){
		cin>>year>>after;
		// 找到第一个闰年
		while(!((year%4==0 && year%100!=0) || year%400==0)){
			year += 1;
		}
		k = 1;
		while(k < after){
			year += 4;
			if((year%4==0 && year%100!=0) || year%400==0){
				k++;
			}
		}
		cout<<year<<endl;
	}
	return 0;
}

// 1.2.4 find your present (2)   异或功能不简单，我也是看别人文章才知道的
//关键在于这里其他的数字都是出现的次数都是偶数次，可以通过异或抵消掉
//我顺便还看了两个数不通过第三个数进行相互交换的异或方法，复习了一下异或
#include<iostream>
using namespace std;
int main()
{
	int n;
	while(cin>>n){
//		cout<<"开始"<<endl;
		if(n == 0){
//			cout<<"停止"<<endl;
			break;
		}else{
			int s = 0;
			int k = n;
			while(k--){
				int i;
				cin>>i;
				s^=i;
			}
			cout<<s<<endl;
//			cout<<"结束"<<endl; 
		}
	}
	return 0;
}

//1.2.5 The Seven Percent Solution
#include<iostream>
#include<string>
using namespace std;
int main()
{
	string s;
	while(getline(cin, s)){
		if(s[0] == '#'){
			break;
		}
		int len = s.length();
		for(int i = 0; i < len; i++){
			switch(s[i]){
				case ' ': cout<<"%20";break;
				case '!': cout<<"%21";break;
				case '$': cout<<"%24";break;
				case '%': cout<<"%25";break;
				case '(': cout<<"%28";break;
				case ')': cout<<"%29";break;
				case '*': cout<<"%2a";break;
				default: cout<<s[i];
			}
		}
		cout<<endl;
	}
	return 0;
}

//1.2.6 Quicksum 有意思
#include<iostream>
#include<string>
using namespace std;
int main()
{
	string s;
	int len, quicksum;
	while(getline(cin, s)){
		if(s[0] == '#'){
			break;
		}
		len = s.length();
		quicksum = 0;
		for(int i = 0; i < len; i++){
			if(s[i] != ' '){
				int k = s[i] - 64;
				quicksum += k * (i+1);
			}
		}
		cout<<quicksum<<endl;
	}
	return 0;
}

//1.2.7 decimal system 可以的
#include<iostream>
#include<string>
using namespace std;
int main()
{
	int n, r, sum, total;
	string s;
	while(cin>>n){
		getchar();
		if(n == 0){
			break;
		}
		total = 0;
		while(n--){
			getline(cin, s);
			int start = s.find('(');
			int end = s.find(')');
			if(end-start > 2){
				r = 10;
			}else{
				r = s[start+1] - 48;
			}
			sum = 0;
			for(int i = 0; i < start; i++){
				sum = (s[i] - 48) + sum * r;
//				cout<<sum<<endl;
			}
			total += sum;
		}
		cout<<total<<endl;
	}
	return 0;
}

//1.2.8 Vowel Counting
#include<string>
#include<stdio.h>
#include<iostream>
using namespace std;
int main(){
	int n, len, k;
	cin >> n;
	getchar(); // because '\n' also is a string type, so need a getchar here 
	string s;
	while(n--){
		getline(cin, s);
		int len = s.length();
		for(int i=0; i <= len; i++){
			switch(s[i]){
				case 'A':
				case 'E':
				case 'I':
				case 'O':
				case 'U':break;
				case 'a':
				case 'e':
				case 'i':
				case 'o':
				case 'u':s[i] = char(s[i] - 32); break;
				default :
					if(s[i] - 0 <= 90){
						s[i] = char(s[i] + 32);
					}
			}
		}
		cout<<s<<endl;
	}
	return 0;
}