日韩性视频-久久久蜜桃-www中文字幕-在线中文字幕av-亚洲欧美一区二区三区四区-撸久久-香蕉视频一区-久久无码精品丰满人妻-国产高潮av-激情福利社-日韩av网址大全-国产精品久久999-日本五十路在线-性欧美在线-久久99精品波多结衣一区-男女午夜免费视频-黑人极品ⅴideos精品欧美棵-人人妻人人澡人人爽精品欧美一区-日韩一区在线看-欧美a级在线免费观看

歡迎訪問 生活随笔!

生活随笔

當前位置: 首頁 > 编程资源 > 编程问答 >内容正文

编程问答

01-复杂度2 Maximum Subsequence Sum (25 分)

發布時間:2023/11/30 编程问答 47 豆豆
生活随笔 收集整理的這篇文章主要介紹了 01-复杂度2 Maximum Subsequence Sum (25 分) 小編覺得挺不錯的,現在分享給大家,幫大家做個參考.

Given a sequence of?K?integers {?N?1??,?N?2??, ...,?N?K???}. A continuous subsequence is defined to be {?N?i??,?N?i+1??, ...,?N?j???} where?1. The Maximum Subsequence is the continuous subsequence which has the largest sum of its elements. For example, given sequence { -2, 11, -4, 13, -5, -2 }, its maximum subsequence is { 11, -4, 13 } with the largest sum being 20.

Now you are supposed to find the largest sum, together with the first and the last numbers of the maximum subsequence.

Input Specification:

Each input file contains one test case. Each case occupies two lines. The first line contains a positive integer?K?(≤). The second line contains?K?numbers, separated by a space.

Output Specification:

For each test case, output in one line the largest sum, together with the first and the last numbers of the maximum subsequence. The numbers must be separated by one space, but there must be no extra space at the end of a line. In case that the maximum subsequence is not unique, output the one with the smallest indices?i?and?j?(as shown by the sample case). If all the?K?numbers are negative, then its maximum sum is defined to be 0, and you are supposed to output the first and the last numbers of the whole sequence.

Sample Input:

10 -10 1 2 3 4 -5 -23 3 7 -21

Sample Output:

10 1 4 #include<cstdio> const int maxn = 100100; int a[maxn] = {0}; int dp[maxn] = {0}; int s[maxn] = {0};int main(){int n;scanf("%d",&n);bool flag = false;for(int i = 0; i < n; i++){scanf("%d",&a[i]);if(a[i] >= 0) flag = true;}if(!flag){printf("0 %d %d",a[0],a[n-1]);return 0;}//scanf("%d",&n);dp[0] = a[0];for(int i = 1; i < n; i++){if(dp[i-1] + a[i] >= a[i]){dp[i] = dp[i-1]+a[i];s[i] = s[i-1];}else{dp[i] = a[i];s[i] = i;}}int max = dp[0];int k = 0;for(int i = 1; i < n; i++){if(dp[i] > max){max = dp[i];k = i;}}printf("%d %d %d",max,a[s[k]],a[k]);return 0; }

?

轉載于:https://www.cnblogs.com/wanghao-boke/p/10548671.html

總結

以上是生活随笔為你收集整理的01-复杂度2 Maximum Subsequence Sum (25 分)的全部內容,希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯,歡迎將生活随笔推薦給好友。