hdu1003 Max Sum 最大子列和问题

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 208994    Accepted Submission(s): 48944

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
   
   
   
   

用一个ThisSum存放当前子列和，MaxSum为最终的最大子列和，当累加后ThisSum < 0时，ThisSum重置为0，因为当前ThisSum为负数的时候显然对后面的

累加和不会做正的贡献，重新计算新的子列和

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

int a[100000 + 10];

int main()
{
	int T, N;
	scanf("%d", &T);
	for (int t = 1; t <= T; t++) {
		scanf("%d", &N);
		for (int i = 1; i <= N; i++) {
			scanf("%d", &a[i]);
		}
		//s用来存储临时子列起点，fs是最终子列的起点
		//e是最终子列的终点
		int ThisSum, MaxSum, s, e, fs;
		bool pfu;
		pfu = true;
		ThisSum = 0;
		MaxSum = a[1];
		fs = e = 1;
		for (int i = 1; i <= N; i++) {
			//pfu为true说明ThisSum刚刚被置为0，重新记录起点
			if (pfu) s = i;
			pfu = false;
			ThisSum += a[i];
			if (ThisSum > MaxSum) {
				MaxSum = ThisSum;
				fs = s;
				e = i;
			}
			if (ThisSum < 0) {
				pfu = true;
				ThisSum = 0;
			}
		}
		printf("Case %d:\n", t);
		printf("%d %d %d\n", MaxSum, fs, e);
		if (t != T) puts("");
	}
	return 0;
}