HDU 1003 Max Sum

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 172698    Accepted Submission(s): 40230

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

 

我的代码老是超时，估计也是因为开了数组的缘故。

下面是两篇不一样的，我的超时了的，还有的是一个大神的，两个思路不一样。

以下是我的

#include <stdio.h>#include <string.h>int main(){    int T, N;    int f[100001];    int g[100001];    scanf("%d", &T);    int count=0;    while(T--)    {        count++;        scanf("%d", &N);        for (int i=0; i<N; i++)        {            scanf("%d", g+i);        }        int max = g[0];        int s,e;        s = e = 0;        for (int i=0; i<N; i++)        {            f[0]=g[i];            for (int j=i+1; j<N; j++)            {                f[j] = f[j-1]+g[j];                                if (f[j] > max)                {                    s = i; e = j;                    max = f[j];                }                       }        }        printf("Case %d:\n", count);        printf("%d %d %d\n\n", max, s+1, e+1);    }             return 0;}

大神的

 #include <iostream>//http://blog.csdn.net/akof1314/article/details/4757021using namespace std;int main(){int t,n,temp,pos1,pos2,max,now,x,i,j;cin>>t;for (i=1;i<=t;i++){cin>>n>>temp;now=max=temp;pos1=pos2=x=1;for (j=2;j<=n;j++){cin>>temp;//以下判断可以写为 if (now<0)//所以，如果前面的总和为负，那就从当前的重新开始            //对于-1, -2, -3的情况，由于前面max设为了-1，所以无错            //对于-2, -1, -3的情况，由于总和为负所以到temp为-1时会重新开始，            //且now>max，max=-1，所以无错            //对于-3, -2, -1的情况，同理。            //其他更普遍的情况，只要前面总和为负了，就重新开始，且与max比较大小            //否则，只要前面总和为正或零，就一直加，且在加的过程中把期间最大的值            //赋给max（如果有的话） if (now+temp<temp)now=temp,x=j;elsenow+=temp;if (now>max)max=now,pos1=x,pos2=j;}cout<<"Case "<<i<<":"<<endl<<max<<" "<<pos1<<" "<<pos2<<endl;if (i!=t)cout<<endl;}return 0;}

下面是两篇不一样的，我的超时了的，还有的是一个大神的，两个思路不一样。