HDU 1003 Max Sum 最长连续子序列和

最新推荐文章于 2022-05-29 08:47:00 发布

TaoSama

最新推荐文章于 2022-05-29 08:47:00 发布

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HDU 1003 Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 158827 Accepted Submission(s): 37157

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

- - 和前面那道题一样注意两个case之间才多换一行

AC代码如下：

//
//  HDU 1003 Max Sum
//
//  Created by TaoSama on 2015-02-06
//  Copyright (c) 2014 TaoSama. All rights reserved.
//
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
#define CLR(x) memset(x, 0, sizeof(x))

using namespace std;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const int N = 1e5 + 10;

int n;

int main() {
#ifdef LOCAL
	freopen("in.txt", "r", stdin);
//	freopen("out.txt","w",stdout);
#endif
	ios_base::sync_with_stdio(0);
	int t; cin >> t;
	for(int kase = 1; kase <= t; ++kase) {
		cin >> n;
		int ans = -INF, sum = 0, x, l, r, st = 1;
		for(int i=1;i<=n;++i){
			cin>>x;
			sum+=x;
			if(sum > ans){
				ans = sum;
				l = st;
				r = i;
			}
			if(sum < 0) sum = 0, st = i + 1;
		}
		if(kase > 1) cout<<endl;
		cout<<"Case "<<kase<<":"<<endl;
		cout<<ans<<' '<<l<<' '<<r<<endl;
	}
	return 0;
}