HDU 1003 最大连续子段和

最新推荐文章于 2018-08-06 17:44:09 发布

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本文介绍了一个经典的编程问题——最大连续子段和问题，并提供了一种使用动态规划解决该问题的方法。通过输入一系列整数，算法能够找到具有最大和的连续子序列及其起始和结束位置。

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最大连续子段和
HDU 1003
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

//动态规划
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
int dp[100005];//每处的最大子列和
int main(){
	int test;//测试样例数
	int n;//多少个数
	int max,start,end;//三个结果
	int startnow;//当前开始
	int number;
	cin>>test;
	for(int i=1;i<=test;i++){
		cin>>n;
		max=-10000;
		startnow=1;
		start=end=1;
		memset(dp,0,sizeof(dp));
		for(int j=1;j<=n;j++){
			cin>>number;
			dp[j]=dp[j-1]+number;
			if(dp[j]>max){
				max=dp[j];
				start=startnow;
				end=j;
			}
			if(dp[j]<0){
				dp[j]=0;
				startnow=j+1;
			}
		}
		cout<<"Case "<<i<<":\n";
		cout<<max<<' '<<start<<' '<<end<<'\n';
		if(i<test)
			cout<<'\n';
	}
	return 0;
}