hdu Max Sum

点击打开链接 密码：syuct

Max Sum

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 1   Accepted Submission(s) : 1

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Problem Description

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

求最长连续子序列和，并输出起始位置和末位置。

#include<iostream>#include<cstdio>#include<cstring>using namespace std;const int N=1e5+10;int a[N];int inf=1<<29;int main(){    int T,n;    scanf("%d",&T);    int ca=1;    while(T--)    {        scanf("%d",&n);        int sum=0,ma=-inf;        int start=1,end=1,k=1;        for(int i=1;i<=n;i++)        {            scanf("%d",&a[i]);            sum+=a[i];            if(sum>ma)            {                ma=sum;                start=k;                end=i;            }            if(sum<0)            {                sum=0;                k=i+1;            }        }        printf("Case %d:\n",ca++);        printf("%d %d %d\n",ma,start,end);        if(T!=0)            printf("\n");    }    return 0;}