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  • HDU 1024 Max Sum Plus Plus (最大和子序列增强版:求规定下标内的最大值)

    Max Sum Plus Plus

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 17906    Accepted Submission(s): 5864


    Problem Description
    Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult problem.

    Given a consecutive number sequence S1, S2, S3, S4 ... Sx, ... Sn (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ Sx ≤ 32767). We define a function sum(i, j) = Si + ... + Sj (1 ≤ i ≤ j ≤ n).

    Now given an integer m (m > 0), your task is to find m pairs of i and j which make sum(i1, j1) + sum(i2, j2) + sum(i3, j3) + ... + sum(im, jm) maximal (ix ≤ iy ≤ jx or ix ≤ jy ≤ jx is not allowed).

    But I`m lazy, I don't want to write a special-judge module, so you don't have to output m pairs of i and j, just output the maximal summation of sum(ix, jx)(1 ≤ x ≤ m) instead. ^_^
     
    Input
    Each test case will begin with two integers m and n, followed by n integers S1, S2, S3 ... Sn.
    Process to the end of file.
     
    Output
    Output the maximal summation described above in one line.
     
    Sample Input
    1 3 1 2 3
    2 6 -1 4 -2 3 -2 3
     
    Sample Output
    6
    8
     
     1 #include<cstdio>
     2 #include<cstring>
     3 #include<algorithm>
     4 #include<iostream>
     5 #define inf 999999999
     6 //using namespace std;
     7 const int MAX = 1000005;
     8 int max(int a,int b)
     9 {
    10     return a>b?a:b;
    11 }
    12 int a[MAX];
    13 int left[MAX];
    14 int right[MAX];
    15 int main()
    16 {
    17     int n,m,i,j;
    18     while(scanf("%d%d",&m,&n)!=EOF)
    19     {
    20         memset(a,0,sizeof(a));
    21         memset(left,0,sizeof(left));
    22         memset(right,0,sizeof(right));
    23         for(i=1;i<=n;i++)
    24             scanf("%d",&a[i]);
    25         long long count;
    26         for(i=1;i<=m;i++)
    27         {
    28             count=-inf;
    29             for(j=i;j<=n;j++)
    30             {
    31                 left[j]=max(left[j-1],right[j-1])+a[j];
    32                 right[j-1]=count;
    33                 if(left[j]>count)
    34                     count=left[j];
    35             }
    36         }
    37         printf("%I64d
    ",count);
    38     }
    39     return 0;
    40 }
    View Code
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  • 原文地址:https://www.cnblogs.com/caterpillarofharvard/p/4234697.html
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