HDU 1003 Max Sum

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 137183    Accepted Submission(s): 31788

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

 

跟上一题一样。

AC代码：

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int a[100005];int dp[100005];int main(){    int t,i,j,n;    scanf("%d",&t);    for(i=1;i<=t;i++)    {        scanf("%d",&n);        for(j=1;j<=n;j++)            scanf("%d",&a[j]);        dp[1]=a[1];        int r=1,l=1,current=1;        int ma=-1;        for(j=2;j<=n;j++)        {            if(dp[j-1]>=0)//确定dp[j]的值                dp[j]=dp[j-1]+a[j];            else            {                dp[j]=a[j];                current=j;//current是个中间变量,保存起点的值            }//先求出dp[j]的大小然后跟ma比较，在判断起点终点，否则是没有意义的，所以先用current保存起点的值，而终点的值就是当前值，更新的时候起点不一定换但终点一定换            if(dp[j]>=ma)//确定起点终点，这个=要有，没有会wa            {                ma=dp[j];                r=j;                l=current;            }        }        printf("Case %d:\n",i);        printf("%d %d %d\n",ma,l,r);        if(i!=t)        printf("\n");    }    return 0;}