Bridging signals Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 961 Accepted Submission(s): 627 Problem Description 'Oh no, they've done it again', cries the chief designer at the Waferland…
Bridging signals Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 1280 Accepted Submission(s): 832 Problem Description 'Oh no, they've done it again', cries the chief designer at the Waferlan…
Bridging signals Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4452 Accepted Submission(s): 2769 Problem Description 'Oh no, they've done it again', cries the chief designer at the Waferlan…
Problem Description 'Oh no, they've done it again', cries the chief designer at the Waferland chip factory. Once more the routing designers have screwed up completely, making the signals on the chip connecting the ports of two functional blocks cross…
Bridging signals Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 9234 Accepted: 5037 Description 'Oh no, they've done it again', cries the chief designer at the Waferland chip factory. Once more the routing designers have screwed up co…
题意: 给一个数字序列,要求找到LIS,输出其长度. 思路: 扫一遍+二分,复杂度O(nlogn),空间复杂度O(n). 具体方法:增加一个数组,用d[i]表示长度为 i 的递增子序列的最后一个元素,且该元素总是保持当前最小.初始化d[1]=A[i],当前LIS的长度len=1.从 2 to n,若A[i]>d[len],则d[++len]=A[i],否则,在数组d中找到A[i]应该插入的位置,代替掉那个第一个比它大的数字,比如d[k]<A[i]<=d[k+1],直接将A[i]代替掉d[…