Max Sum

    Total Submission(s): 183450 Accepted Submission(s): 42790

    Problem 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

    Author Ignatius.L

    题意：求最大子序列和，

hdu1003 Max Sum（dp）

电脑资料

#include<cstdio>using namespace std;int T,n,m;int post1,post2,x;//post1表示序列起点，post2表示序列终点，x表示每次更新的起点int max,now;//max表示最大子序列和，now表示各个子序列的和int i,j;int main (){    scanf (%d,&T);    for (i=1; i<=T; i++)    {        scanf (%d%d,&n,&m);        max = now = m;        post1 = post2 = x = 1;//初始化        for (j=2; j<=n; j++)        {            scanf (%d,&m);            if (now + m < m)//对于每个数，如果该数加上当前序列和比本身还小            {                now = m;//更新区间                x = j;//更新起点            }            else            now += m;//否则把该数加进序列            if (now > max)//如果当前序列和比已有最大序列和大，更新            {                max = now;                post1 = x;//记录新的起点和终点                post2 = j;            }        }        printf (Case %d:,i);        printf (%d %d %d,max, post1, post2);        if (i != T)        printf ();    }    return 0;}</cstdio>