eMFtQj hIPNtw jQMBWr 所以在JS的世界里, 很需要一种侵入性不那么强而且易读的异步回调解决方案, ES7的规范里, ECMAScript又提出了async/await, 使用这个方案仍然需要使用Promise来改写异步过程, 但是需要用then来改写执行过程, 而是可以像同步代码一样将结果直接作为执行结果返回, 可阅读性也非常好. 这个方案的 hYcTaL hctuh jMfvbc gFwJbt [object Object],[object Object],[object Object],[object Object] ikkyVr bPjPUf cofShm lmYuZf euExUQ hDyWCg hUVKYO [object Object] uQPrX ekvxDO bpqJVJ iGaadT iGXldS hOxXax emJCYl ddGdPh iBSbYp dwejLW esHzLO"> /* sc-component-id: sc-bdVaJa */ .sc-bdVaJa {} .jTcPxY{height:100%;overflow:auto;} /* sc-component-id: sc-bwzfXH */ .sc-bwzfXH {} .iYgZUj{color:#BBB;margin:10px 0 20px 0;} /* sc-component-id: sc-htpNat */ .sc-htpNat {} .knYzzU{position:relative;padding:10px 0 10px 0;width:calc(100% - 10px);height:50px;-webkit-transition:-webkit-transform 0.6s;-webkit-transition:transform 0.6s;transition:transform 0.6s;} .knYzzU:hover{background-color:#EEE;-webkit-transform:translate(10px,0);-ms-transform:translate(10px,0);transform:translate(10px,0);}.null{position:relative;padding:10px 0 10px 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Life of xhu

标准库学习 & HOJ_1280 & HOJ_1284

Feb 01, 2014  |  C++  |  STL  |  Algorithm  |  HOJ
  1. C++中的哈希:C++中其实已经提供了哈希表(散列表)的实现,即标准库map,语法如下:

     #include <map>
    
     map<int, string> test;

    这样就建立了一个key为整数,value为字符串的哈希表,添加一个key-value对和获得一个keyvalue也很简单,使用与数组类似的方法即可

     test[1] = "test";
     cout << test[1] << endl;     // => test

    当然,这个标准库也有自己的迭代器,在查找元素和删除元素时都要用到,使用find函数创造迭代器后,可以进行相关的操作,代码如下

     //查找操作,在找不到键值对时,迭代器最终会指向表尾        
     search_test = test.find(3);
     if(search_test == test.end())
       cout << "there is not a key-value pair which key is 3" << endl;
    
     //删除操作,如果迭代器找到该键值对,会停止迭代,这时可以调用删除的函数
     delete_test = test.find(2);
     test.erase(delete_test);   //即删除该迭代器指向的键值对
  2. atoi函数:即将字符串转为整数的函数,包含在stdlib.h中,用法很简单:

     int test = atoi("123");

然后,照理是两道HOJ水题,话说我今天把我之前做的一些网页的CSS也加上了overflow: auto,好看了好多啊有木有~~

HOJ_1280: 简单的字符串处理,理解了题意基本上就能做出来,代码如下:

#include <iostream>
#include <cstring>
#include <string>
using namespace std;

int main(){
    int n, order_len, message_len;
    char order[10], message[30], tmp;
    cin >> n;
    while(n){
        cin >> order;
        order_len = strlen(order);
        cin >> message;
        message_len = strlen(message);
        //cout << order << ' ' << message << endl;
        for(int i = order_len - 1; i >= 0; i--){
            switch(order[i]){
                case 'J':
                    tmp = message[message_len - 1];
                    for(int j = message_len - 1; j > 0; j--)
                        message[j] = message[j - 1];
                    message[0] = tmp;
                    break;
                case 'C':
                    tmp = message[0];
                    for(int j = 0; j < message_len - 1; j++)
                        message[j] = message[j + 1];
                    message[message_len - 1] = tmp;
                    break;
                case 'E':
                    for(int j = message_len / 2 - 1, k = message_len - 1; j >= 0; j--, k--){
                        tmp = message[k];
                        message[k] = message[j];
                        message[j] = tmp;
                    }
                    break;
                case 'A':
                    for(int j = 0, k = message_len - 1; j < k; j++, k--){
                        tmp = message[k];
                        message[k] = message[j];
                        message[j] = tmp;
                    }
                    break;
                case 'P':
                    for(int j = 0; j < message_len; j++){
                        if(message[j] >= '1' && message[j] <= '9')
                            message[j] = (char)((int)message[j] - 1);
                        else if(message[j] == '0')
                            message[j] = '9';
                    }
                    break;
                case 'M':
                    for(int j = 0; j < message_len; j++){
                        if(message[j] >= '0' && message[j] <= '8')
                            message[j] = (char)((int)message[j] + 1);
                        else if(message[j] == '9')
                            message[j] = '0';
                    }
                    break;
                default:
                    break;
            }
        }
        cout << message << endl;
        n--;
    }
}  

HOJ_1284: 这道题本来我一看到就想用后缀表达式来做,但是想了一会儿,觉得用后缀表达式有点难,再想,我去,这题用栈不就行了么,大致思想就是,在遍历开始和左括号开始,都独立出一个栈,而遇到右括号,则将目前栈里的值反悔并作为上一个栈的栈顶元素,遇到数字则与当前栈顶相加,最后将剩下的一个栈里所有元素相加就可以得到结果。

好吧,我觉得我这样说是没人能懂的,但是这道题的确不错,我觉得做这样的栈模拟题很有意思啊,要是能有人给我一个HOJ上的栈模拟题列表就好了。。。

话不多说,代码如下:

#include <iostream>
#include <map>
#include <string>
#include <cstring>
#include <cstdlib>
using namespace std;
#define N 100

typedef struct {
    int data[N], top;
} STACK;

map<string, int> atlist;

int isupper(char ch) {
    return ch >= 'A' && ch <= 'Z';
}

int islower(char ch) {
    return ch >= 'a' && ch <= 'z';
}

int isdigit(char ch) {
    return ch >= '0' && ch <= '9';
}

char cmd[N];

int solve(int &index) {
    STACK s;
    string tmp;
    char ch;
    int ans = 0, i, t;
    s.top = -1;
    while (cmd[index] && cmd[index] != ')') {
        ch = cmd[index];
        if (isdigit(ch)) {
            char num[N];
            i = 0;
            while (isdigit(cmd[index]))
                num[i++] = cmd[index++];
            num[i] = 0;
            t = atoi(num);
            s.data[s.top] *= t;
        } else if (isupper(ch)) {
            index++;
            tmp.clear();
            tmp += ch;
            while (islower(cmd[index]))
                tmp += cmd[index++];
            if (atlist[tmp] == 0)
                return -1;
            s.data[++s.top] = atlist[tmp];
        } else if (ch == '(') {
            t = solve(++index);
            if (t == -1)
                return -1;
            else
                s.data[++s.top] = t;
        }
    }
    index++;
    for (i = 0; i <= s.top; i++)
        ans += s.data[i];
    return ans;
}

int main() {
    while (cin >> cmd) {
        int num;
        if (!strcmp(cmd, "END_OF_FIRST_PART"))
            break;
        cin >> num;
        atlist[string(cmd)] = num;
    }
    while (cin >> cmd) {
        int ans, index = 0;
        if (cmd[0] == '0')
            break;
        ans = solve(index);
        if (ans == -1)
            cout << "UNKNOWN" << endl;
        else
            cout << ans << endl;
    }
    return 0;
}

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