【PAT甲】1007 Maximum Subsequence Sum (25分)，求最大字段和及区间

problem

1007 Maximum Subsequence Sum (25分)
Given a sequence of K integers { N
​1
​​ , N
​2
​​ , …, N
​K
​​ }. A continuous subsequence is defined to be { N
​i
​​ , N
​i+1
​​ , …, N
​j
​​ } where 1≤i≤j≤K. The Maximum Subsequence is the continuous subsequence which has the largest sum of its elements. For example, given sequence { -2, 11, -4, 13, -5, -2 }, its maximum subsequence is { 11, -4, 13 } with the largest sum being 20.

Now you are supposed to find the largest sum, together with the first and the last numbers of the maximum subsequence.

Input Specification:
Each input file contains one test case. Each case occupies two lines. The first line contains a positive integer K (≤10000). The second line contains K numbers, separated by a space.

Output Specification:
For each test case, output in one line the largest sum, together with the first and the last numbers of the maximum subsequence. The numbers must be separated by one space, but there must be no extra space at the end of a line. In case that the maximum subsequence is not unique, output the one with the smallest indices i and j (as shown by the sample case). If all the K numbers are negative, then its maximum sum is defined to be 0, and you are supposed to output the first and the last numbers of the whole sequence.

Sample Input:
10
-10 1 2 3 4 -5 -23 3 7 -21
Sample Output:
10 1 4

• 给定一个长为n的序列
• 求最大子序列和，以及该序列的开始和结束值。

solution

• 求最大和：tmp为到第i个数时（包含第i个数）的最大字段和，转移时直接考虑加上第i+1个数是大了还是小了（更新全局最大字段和sum，已覆盖所有状态），转移后tmp加上第i+1个数继续往下走（因为必须连续），若是负数则直接断开，前面后面分别求。
• 求序列区间：left和right为开始值和结束值，在每次更新tmp和sum的时候更新对应的区间即可。

#include
#include
using namespace std;
int a[10010];
int main(){
     
	int n; cin>>n;
	int sum = -1, tmp = 0, tmpx = 1;
	int left = 1, right = n;
	for(int i = 1; i <= n; i++){
     
		cin>>a[i];
		tmp += a[i];
		if(tmp < 0){
     
			tmp = 0;
			tmpx = i+1;
		}else if(tmp > sum){
     
			sum = tmp;
			left = tmpx;
			right = i;
		}
	}
	if(sum<0)sum = 0;
	cout<<sum<<" "<<a[left]<<" "<<a[right];
	return 0;
}