hdu 1003

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 

题意：

给你n个数a[1],a[2],a[3]...a[n],要你求出最大连续子段和，并且输出这个和，和这个子段的起始位置和终止位置，如果这个子段有多种可能，输出第一种结果即可。

思路：

前段时间做个类似的题目，用的是暴力求解，用的两重循环，但是这个题目要求是n<100000，用两重循环肯定会超时，因此我们必须换一种思路，必须减少循环次数，当输入的时候，累计求和，用pos记录位置，star记录起点，end记录终点，如果sum小于0.就不必往下面累加了。

代码:

#include<cstdio>using namespace std;#define INF 0x7fffffffint const maxn=100000+10;int a[maxn];int main(){    int kcase=1;    int T,n,max,star,end,pos,sum;//max为最大字段和，star为枚举起始位置，end为枚举终止位置，sum为字段和，pos用来更新枚举起始位置    scanf("%d",&T);    while(T--)    {        scanf("%d",&n);        max=-INF;        pos=sum=0;        for(int i=0;i<n;i++)            {                scanf("%d",&a[i]);                sum+=a[i];                if(sum>max)                {                    max=sum;                    star=pos;                    end=i;                }                if(sum<0)                {                    sum=0;                    pos=i+1;                }            }        printf("Case %d:\n",kcase++);        printf("%d %d %d\n",max,star+1,end+1);//数组下标是从0开始的，而数的位置是从1开始的        if(T)          printf("\n");    }    return 0;}