【DP经典系列】最大连续子序列和

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

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5 

 

Sample Output

Case 1:14 1 4Case 2:7 1 6 

题意：仍旧是最大连续子序列和；

注意输出格式就行；

AC代码：

#include <bits/stdc++.h>#define ll long longusing namespace std ;int main(){    int l , r , maxi , temp , now  ;    int p1,  p2 ;    int t ;    cin>> t ;    for(int cas = 1 ; cas <=t ; cas++)    {        int n ;        cin>>n>>temp ;        now=maxi=temp;        r=l=p1=p2=1;        for(int i=2; i<=n; i++)        {            cin>>temp;            if(temp>now+temp)            {                now=temp;                p1=i;            }            else            {                now+=temp;            }            if(now>maxi)            {                l=p1;                r=i;                maxi=now;            }        }        printf("Case %d:\n",cas);        printf("%d %d %d\n",maxi , l , r);        if(cas!=t)            printf("\n");    }    return 0 ;}