最大子串和

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


题目大意：求最大连续子序列和，输出最大的和以及这个子序列的开始值和结束值。如果所有数都小于0，那么认为最大的和为0，并且输出首尾元素。（首尾元素的坐标要尽量小）

int n;
ll a[N],dp[N];
int b[N],c[N];
multiset<P>se;
multiset<P>::iterator  it;
//dp[i] : 以a[i]结尾的最大子串和
int  main()
{
    
    scanf("%d",&n);
    bool flag=0;
    gep(i,1,n)
    {
    scanf("%lld",&a[i]);
    if(a[i]>=0) flag=1;
    }
    if(!flag){
        printf("0 %lld %lld
",a[1],a[n]);
        return 0;
    }
    gep(i,1,n){
        b[i]=1;
        c[i]=i;
    }
    dp[1]=a[1];
    ll MAX=dp[1];
    gep(i,2,n){
        if(dp[i-1]+a[i]>dp[i]){
            dp[i]=dp[i-1]+a[i];
            b[i]=b[i-1];
        }
        else{
            dp[i]=a[i];
            b[i]=i;
        }
        MAX=max(MAX,dp[i]);
    }
    gep(i,1,n){
        if(dp[i]==MAX){
            se.insert(P(b[i],c[i]));
        }
    }
    it=se.begin();
    int x=(*it).first;
    int y=(*it).second;
    printf("%lld %lld %lld
",MAX,a[x],a[y]);    //要保证 x,y小
    return 0;
}