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HDOJ1003:Max Sum——简单动态规划

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 DP问题。注意数组必须要开到100000以上才可以。 思路大体是开一个临时的子串并不断跟sum进行比较。复习一下动态规划再看。 代码:

#include
#include
#include
#include
#include
#include
using namespace std;
int n, s, e;
int a[100002];

int maxsub(int a[])
{
	s = e = 1;
	int sum, j, b, bs, be;
	sum = b = a[0];
	for (j = 0; j= 0 && j != 0)
		{
			b = b + a[j];
			be = j + 1;
		}
		else
		{
			b = a[j];
			bs = j + 1;
			be = j + 1;
		}
		if (b>sum)
		{
			sum = b;
			s = bs;
			e = be;
		}
	}
	return sum;
}

int main()
{
	int t, i, k = 1, sum;
	scanf(“%d”, &t);
	while (t–)
	{
		scanf(“%d”, &n);
		for (i = 0; i

另一版本

#include 
using namespace std;
int *arr, first, last, sum;
void MaxSub(int *arr, int count, int &first, int &last, int ∑)
{
	//tFirst,tSum为临时的最大子序列的第一个元素与和
	int tFirst, tSum, i;
	//初始化
	first = tFirst = last = 0;
	sum = tSum = arr[0];
	//遍历该数组
	for (i = 1; i= 0)
			tSum += arr[i];
		else
		{
			tSum = arr[i];
			tFirst = i;
		}
		//如果临时的最大值大于实际的最大值,则更新子序列的first, last, sum
		if (sum < tSum)
		{
			sum = tSum;
			first = tFirst;
			last = i;
		}
	}
}
int main()
{
	int num, n, i, j;
	bool top = true;
	cin >> num;
	for (j = 0; j> n;
		arr = new int[n];
		for (i = 0; i> arr[i];
		MaxSub(arr, n, first, last, sum);
		if (top)
			top = false;
		else
			cout << endl;
		cout << "Case " << j + 1 << ":" << endl
			<< sum << " " << first + 1 << " "
			<< last + 1 << endl;
		delete[] arr;
	}
	return 0;
}
//求最大子序列之和,arr是原来的数组,count是数组的元素个数,
//first,last是所求子序列的第一个元素,最后一个元素的下标,sum是所求子序列的元素之和

部分参考自网络。


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