HDU-1003 Max Sum

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 252943    Accepted Submission(s): 60001

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

 

题目的意思是求一段连续的和最大序列，如果有多个序列，则输出第一个。

代码思想用当前部分和与之前的最大部分和比较，若比之前的大，则更新。

若当前部分和小于零，即可以废弃当前部分和，寻求下一个。

最后判断全是负数的情况

 

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;

int main()
{
    //freopen("in.txt","r",stdin);
    int T, n, subsum, ans;              //subsum：当前部分和，ans：答案
    int N = 1;
    int Sta, End, Sta1;            //Sta：至今最大和起始位置，End：至今最大和终止位置，Sta1当前部分和的起始位置
    int a[110000];
    scanf("%d",&T);
    while( T--) {
        scanf("%d",&n);
        if(N != 1) printf("\n");
        printf("Case %d:\n",N++);
        ans = -1, Sta = End = Sta1 = 1, subsum = 0;
        for(int i = 1 ; i <= n ; i++ ) {
            scanf("%d",&a[i]);
            subsum += a[i];
            if(subsum < 0) {Sta1 = i+1 ; subsum = 0;}//当前部分和是负数，则遗弃当前段部分和，寻找下一段
            else {
                if(subsum > ans) {//当前部分和大于之前的某一段部分和，则更新
                        Sta = Sta1;
                        End = i;
                        ans = subsum ;
                }
            }

        }
        if(ans == -1) {//判断全是负数的情况
            ans = -100000;
            for(int i = 1; i <= n; i++) {
                if(ans < a[i]) {
                    ans = a[i];
                    Sta = End = i;
                }
            }
            printf("%d %d %d\n",ans ,Sta, End);
        }
        else
            printf("%d %d %d\n",ans, Sta, End);
     }
    return 0;
}