洛谷 1115 最大子段和、HDU 1003 Max Sum（最大字段和问题）-白红宇

洛谷 1115 最大子段和、HDU 1003 Max Sum（最大字段和问题）

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发布时间：2019-03-04

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洛谷 1115 最大子段和

题目描述

给出一段序列，选出其中连续且非空的一段使得这段和最大。

输入输出格式

输入格式：

第一行是一个正整数N，表示了序列的长度。

第二行包含N个绝对值不大于10000的整数Ai，描述了这段序列。

输出格式：

一个整数，为最大的子段和是多少。子段的最小长度为1。

#include 
   
    #include 
    
     using namespace std;int n, maxx, sum, t;int main(){   	cin>>n;	cin>>t;	maxx=t;	sum=t;	for (int i=2; i<=n; i++){   		cin>>t;		sum=sum>0?sum:0;		sum+=t;		maxx=sum>maxx?sum:maxx;	}	printf("%d", maxx);	return 0;}

HDU 1003 Max Sum

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

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

#include 
   
    #include 
    
     #include 
     
      #include 
      
       #include 
       
        using namespace std;int main(){   	int T, n;	scanf("%d", &T);	for (int k=1; k<=T; k++){   		scanf("%d", &n);		int t;		scanf("%d", &t);		int begin=1, end=1, sum=t, maxx=t, begint=1, endt=1;		for (int i=2; i<=n; i++){   			scanf("%d", &t);			if (sum<0){   				sum=0;				begint=i;				endt=i;			}			sum+=t;			if (sum>maxx){   				endt=i;				begin=begint;				end=endt;				maxx=sum;			}		}		printf("Case %d:\n", k);		printf("%d %d %d\n", maxx, begin, end);		if (k!=T)			printf("\n");	}	return 0;}