ACM 之 J - Max Sum

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

理解

给你一系列的数，找出一段数，使得这段数的和最大，输出最大和和这段数的始末。

代码部分

#include 
using namespace std;
int t,n,l,r,a,w;
long long sum,max;
int main(){
    scanf("%d",&t);getchar();
    for(int i=1;i<=t;i++){
        scanf("%d",&n);getchar();
        l=1;r=0;w=1;
        sum=0;max=-100000000;
        for(int j=1;j<=n;j++){
            scanf("%d",&a);
            sum+=a;
            if(sum>max){
                max=sum;
                r=j;
                l=w;
            }
            if(sum<0){
                sum=0;
                if(a<0)
                    w=j+1;
                else
                    w=j;
            }
        }getchar();
        printf("Case %d:\n%lld %d %d\n",i,max,l,r);
        if(i!=t) printf("\n");
    }
    return 0;
}

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