HDU 1003 Max sum

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 159198    Accepted Submission(s): 37255

Problem Description

Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.

 

Input

The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).

 

Output

For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.

 

Sample Input

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

 

Sample Output

Case 1:14 1 4Case 2:7 1 6

 

Author

Ignatius.L

 

解题思路：

这道题，看起来很水，也就是个dp求最大连续子序列的和。。那么只要定义好状态就可以了，dp[i] 表示的是以a[i]为结尾的最大子序列的和。

那么状态转移方程为：dp[i] = max( dp[i-1]+a[i],a[i] )

不说了，直接上代码：

# include<cstdio># include<iostream>using namespace std;# define MAX 100000int mat[MAX];struct node{    int l;    int r;}seg[MAX];int main(void){    int n;    int icase = 0;    int t;cin>>t;    while ( t-- )    {        icase++;        cin>>n;        for ( int i = 1;i <= n;i++ )        {            cin>>mat[i];        }        seg[1].l = 1;        seg[1].r = 1;        for ( int i = 2;i <= n;i++ )        {            if ( mat[i] <= mat[i-1]+mat[i] )            {                mat[i] = mat[i-1]+mat[i];                seg[i].l = seg[i-1].l;                seg[i].r = i;            }            else            {                seg[i].l = seg[i].r = i;            }        }        int maxID = 1;        int _max = mat[1];        for ( int i = 2;i <= n;i++ )        {            if ( mat[i] > _max )            {                _max = mat[i];                maxID = i;            }        }        printf("Case %d:\n",icase);        printf("%d %d %d\n",_max,seg[maxID].l,seg[maxID].r);        if ( t != 0 )            cout<<endl;    }    return 0;}