杭電HDOJ 1003 解題報告

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 103698    Accepted Submission(s): 23844

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

2 5 6 -1 5 4 -7 7 0 6 -1 1 -6 7 -5

 

Sample Output

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

 

Author

Ignatius.L

這是一道簡單的DP題。

解題思路：

題意：尋找所給數組的最大連續子序列和，並紀錄始末位置。

如果從i到j（i《＝j）的和爲正則繼續，並一直紀錄最大值；爲負則從j＋1重新開始累加開始。同時紀錄始末位置。

注意格式。其他就不多說了。

代碼1、

#include<stdio.h>

main()
{
    int i,j=0,n,m,s,t,l,r,w;
    int a[100000],b[2];

    scanf("%d",&n);
    w=n;
    while(n--)
    {
        s=0;r=0;l=0;b[0]=0;b[1]=0;
        scanf("%d",&m);
        for(i=0;i<m;i++)
        {
            scanf("%d",&a[i]);
            if(i==0) t=a[i];
            s+=a[i];
            if(s>t) {
                t=s;
                if(i>r)  r=i;
                b[0]=l;b[1]=r;
            }
            if(s<0) {s=0;l=i+1;r=0;}
        }
        printf("Case %d:\n",++j);
        printf("%d %d %d\n",t,b[0]+1,b[1]+1);
        if(j<w)  printf("\n");
    }
}

代碼2、

（比前面的更簡練，待續）

測試數據：

4 0 0 2 0 —— 2 1 3

6 2 7 -9 5 4 3 —— 12 1 6

4 0 0 -1 0 —— 0 1 1

7 -1 -2 -3 -2 -5 -1 -2 —— -1 1 1

6 -1 -2 -3 1 2 3 —— 6 4 6

5 -3 -2 -1 -2 -3 —— -1 3 3