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

解析：dp思想

每次往后遍歷，每加一個數，temp+a[i]，若temp<0,說明是加的是復數，則舍棄前面的，因為負數永遠會拉低temp，更新tempindex（遍歷的位置）。

若temp>sum，則說明數是正數，更新sum=temp，st=tempindex，en=i。

#include<bits/stdc++.h> using namespace std;#define e exp(1) #define pi acos(-1) #define mod 1000000007 #define inf 0x3f3f3f3f #define ll long long #define ull unsigned long long #define mem(a,b) memset(a,b,sizeof(a)) int gcd(int a,int b){return b?gcd(b,a%b):a;}const int maxn=1e4+5; int a[maxn];int main() {int n;scanf("%d",&n);int sum=-1,temp=0,tempindex=0,st=0,en=n-1;for(int i=0; i<n; i++){scanf("%d",&a[i]);temp+=a[i];if(temp<0){temp=0;tempindex=i+1;}else if(temp>sum){sum=temp;st=tempindex;en=i;}}if(sum<0)sum=0;printf("%d %d %d\n",sum,a[st],a[en]);return 0; }

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