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

HOJ_1224 & HOJ_1247

Jan 26, 2014  |  HOJ  |  C++  |  Algorithm

昨天因为要给可儿讲睡前故事,所以没有刷水题,今天继续:

  • 1224: 相当水的一道题,就是考冒泡排序的次数,不过题目中相当于是一个循环序列,分析示例可以发现,其实就是相当于把这个循环的序列在中间分成两部分,然后对两部分进行冒泡排序,比如1 2 3 4,可以分成1 23 41 2 3 4 5可以分成1 23 4 5`,来分别进行排序,再把次数相加即可得到答案。

    代码如下:

      #include <iostream>
      using namespace std;
    
      int main(){
          int amount, n;
          cin >> amount;
          while(amount){
              cin >> n;
              if(n % 2 == 0)
                  cout << n / 2 * (n / 2 - 1) << endl;
              else
                  cout << n / 2 * (n / 2 - 1) / 2 + n / 2 * (n / 2 + 1) / 2 << endl;
              amount--;
          }
      }
  • 1247: 我发现ACM很喜欢在质数啊、互质、因数啊这些概念上出题啊。。。
    这道题如果真的一个个去求约数然后求和,结果想都不用想肯定是超时,我感觉没有什么思路,点进Discuss,发现一个大牛给了一个方法,用筛法,直接打表给答案,筛法,感觉很高端的样子啊,立刻Google之,发现原来是一种求质数的方法,在百度百科上找到了解释:

    先把N个自然数按次序排列起来。1不是质数,也不是合数,要划去。第二个数2是质数留下来,而把2后面所有能被2整除的数都划去。2后面第一个没划去的数是3,把3留下,再把3后面所有能被3整除的数都划去。3后面第一个没划去的数是5,把5留下,再把5后面所有能被5整除的数都划去。这样一直做下去,就会把不超过N的全部合数都筛掉,留下的就是不超过N的全部质数。

    而这题使用筛法,思路就是,建立一个全为0的数组num[n],然后从1开始遍历数组,对每一个项num[i],其之后所有角标能被i整除的项,值都加上i。比如遍历到2,就把num[4], num[6], num[8]...的值加2,这样,数组中任意一项,其角标对应的约数和就是这一项的值。然后有输入时,直接查数组输出即可。

    代码如下:

      #include <iostream>
      #include <iomanip>
      using namespace std;
    
      int main(){
          int num[60001], tmp;
          for(int i = 0; i < 60001; i++)
              num[i] = 0;
          for(int i = 1; i <= 60000; i++){
              for(int j = i * 2; j <= 60000; j += i)
                  num[j] += i;
          }
          cout << "PERFECTION OUTPUT" << endl;
          while(cin >> tmp){
              if(tmp == 0){
                  cout << "END OF OUTPUT" << endl;
                  break;
              }else{
                  cout << setw(5) << tmp << "  ";
                  if(num[tmp] == tmp)
                      cout << "PERFECT" << endl;
                  else if(num[tmp] > tmp)
                      cout << "ABUNDANT" << endl;
                  else
                      cout << "DEFICIENT" << endl;
              }
          }
      }

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