HDU 1003

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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

考虑一个“对于前n个数最大的必须到达末尾的连续子序列”，我就把简称为MESS（Max end sub-sequence）

例如“6 -1 5 4 -7”这个数列，对于前3个数“6 -1 5”,它从末尾出发，设其和最大的子序列为MESS[3]，

那么对于前四个“6 -1 5 4”，MESS[4]= ( (MESS[3]>0) ? (MESS[3]+num[4]) : (num[4]) );

另外，最后遍历MESS数组，找到最大的那个值就是answer，那个位置就是子序列的末尾位置，另外为了记录子序列的开始位置，我把MESS定义成了结构体……里面定义一个sum用来存最大的和，另外定义一个begin来储存开始位置。

#include<cstdio>int num[100003];typedef struct{int sum;int begin;}type;type MESS[100003];int main(){int t,n;scanf("%d",&t);for(int kase=1;kase<=t;kase++){scanf("%d",&n);for(int i=1;i<=n;i++) scanf("%d",&num[i]);MESS[1].sum=num[1];MESS[1].begin=1;for(int i=2;i<=n;i++){if(MESS[i-1].sum<0){MESS[i].sum=num[i];MESS[i].begin=i;}else{MESS[i].sum=MESS[i-1].sum+num[i];MESS[i].begin=MESS[i-1].begin;}}type max=MESS[1];int end=1;for(int i=2;i<=n;i++) if(max.sum<=MESS[i].sum) {max=MESS[i];end=i;}if(kase>=2) printf("\n");printf("Case %d:\n%d %d %d\n",kase,max.sum,max.begin,end);}}

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