牛客oj 习题4.4浮点数加法（大数相加）&&习题4.5后缀子串排序（Map）

太长时间不写大数了，练大数是练习字符串操作最好的方法。这种题最重要的就是细心。

我的思路是先把大数字拆成整数部分和小数部分，然后将整数部分和小数部分按位数多少找出长串和短串，然后把短串加到长串上去，用进位符carry模拟进位操作。最后再把小数部分删掉。总体来说挺麻烦的，比赛或考试时候用java就可以了。

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <string>
#include <cstring>
#include <vector>
#include <map>
#include <cctype>
#include <climits>

using namespace std;

const int MAXN = 105;
const int INF = INT_MAX;

string str1, str2;
string int1, decimal1, int2, decimal2;

void exchange(string &x, string &y){
	string tmp = x;
	x = y;
	y = tmp;
}

int main(){
//	freopen("in.txt", "r", stdin);
	bool flag, carry;
	int cur;
	while(getline(cin, str1)){
		getline(cin, str2);
		int length1 = str1.size(), length2 = str2.size();
		//第一行拆开 
		flag = true;
		cur = 0;
		for(int i = 0; i < length1; i++){
			if(str1[i] == '.'){
				flag = false;
				cur = 0;
				continue;
			}
			if(flag){
				int1.insert(cur, 1, str1[i]);
				cur++;
			}
			else{
				decimal1.insert(cur, 1, str1[i]);
				cur++;
			}
		}
		//第二行拆开 
		flag = true;
		cur = 0;
		for(int i = 0; i < length2; i++){
			if(str2[i] == '.'){
				flag = false;
				cur = 0;
				continue;
			}
			if(flag){
				int2.insert(cur, 1, str2[i]);
				cur++;
			}
			else{
				decimal2.insert(cur, 1, str2[i]);
				cur++;
			}
		}
		//保证编号1为最大
		if(int1.size() < int2.size()) exchange(int1, int2);
		if(decimal1.size() < decimal2.size()) exchange(decimal1, decimal2);
		//小数部分相加
		carry = false;
		for(int i = decimal2.size()-1; i >= 0; i--){
			if(carry){//有进位 
				if((decimal1[i]-'0')+(decimal2[i]-'0')+1 >= 10){
					decimal1[i] = (((decimal1[i]-'0')+(decimal2[i]-'0')+1) % 10) + '0';
					carry = true;
				}
				else{
					decimal1[i] = ((decimal1[i]-'0')+(decimal2[i]-'0')+1) + '0';
					carry = false;
				}
			}
			else{//无进位 
				if((decimal1[i]-'0')+(decimal2[i]-'0') >= 10){
					decimal1[i] = (((decimal1[i]-'0')+(decimal2[i]-'0')) % 10) + '0';
					carry = true;
				}
				else{
					decimal1[i] = (decimal1[i]-'0')+(decimal2[i]-'0') + '0';
					carry = false;
				}
			}
		} 
		//整数部分相加
		int end = (int1.size()-int2.size());
		int int2pos = int2.size()-1;
		for(int i = int1.size()-1; i >= end; i--){
			if(carry){//有进位 
				if((int1[i]-'0')+(int2[int2pos]-'0')+1 >= 10){
					int1[i] = (((int1[i]-'0')+(int2[int2pos]-'0')+1) % 10) + '0';
					carry = true;
				}
				else{
					int1[i] = ((int1[i]-'0')+(int2[int2pos]-'0')+1) + '0';
					carry = false;
				}
			}
			else{//无进位 
				if((int1[i]-'0')+(int2[int2pos]-'0') >= 10){
					int1[i] = (((int1[i]-'0')+(int2[int2pos]-'0')) % 10) + '0';
					carry = true;
				}
				else{
					int1[i] = ((int1[i]-'0')+(int2[int2pos]-'0')) + '0';
					carry = false;
				}
			}
			int2pos--;
		} 
		//剩余部分 
		int start = (int1.size()-int2.size()-1);
		for(int i = start; i >= 0; i--){
			if(carry){
				if((int1[i]-'0')+1 >= 10){
					int1[i] = (((int1[i]-'0')+1) % 10) + '0';
					carry = true;
				}
				else{
					int1[i] = ((int1[i]-'0')+1) + '0';
					carry = false;
				}
			}
			else break;
		}
		if(carry) int1.insert(0, 1, '1');
		for(int i = decimal1.size()-1; i >= 0; i--){
			if(decimal1[i] != '0') break;
			decimal1.erase(i);
		}
		cout << int1 << "." << decimal1 << endl;
		int1.clear();
		int2.clear();
		decimal1.clear();
		decimal2.clear();
	}
	return 0;
}

遇到字符串排序，直接利用map的红黑树功能即可，第二个参数只是摆设。。

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <string>
#include <cstring>
#include <vector>
#include <map>
#include <climits>

using namespace std;

const int MAXN = 105;
const int INF = INT_MAX;

map<string, int> table;
string str, tmp;

int main(){
//	freopen("in.txt", "r", stdin);
	while(getline(cin, str)){
		int length = str.size();
		for(int i = 0; i < length; i++){
			tmp = str.substr(i, length-i);
			table.insert(pair<string, int>(tmp, i));
		}
		map<string,int>::iterator it;
		for(it = table.begin(); it != table.end(); it++){
			cout << it->first << endl;
		}
		str.clear();
		table.clear();
	}
	return 0;
}