【HDOJ 1003】 Max Sum

Problem Description

Given a sequence a[1],a[2],a[3]......a
, 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

2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5

Sample Output

Case 1:
14 1 4

Case 2:
7 1 6

解题思路：

求最大子串和。状态转移方程为：s[i].sum = (s[i-1].sum < 0) ? num[i] : sum[i-1].sum + num[i]； （s[i].sum为 以第i个位置为结尾的所有子串中最大的和）最后对s[i].sum(i = 1,...,n)排序求最大值就行了。用到结构体，存有sum,start,end等信息。

参考代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 100010;
struct P{
int start;
int end;
int sum;
}s[maxn];
int num[maxn];
bool cmp(const P& p1,const P& p2){
return p1.sum > p2.sum;
}
int main()
{
int T,n,i,k;
scanf("%d",&T);
for (k = 1;k <= T;k++){
memset(s,0,sizeof(s));
scanf("%d",&n);
for (i = 1;i <= n;i++)
scanf("%d",&num[i]);
s[1].start = s[1].end = 1;
s[1].sum = num[1];
for (i = 2;i <= n;i++){
if(s[i-1].sum < 0){
s[i].sum = num[i];
s[i].start = s[i].end = i;
}else{
s[i].sum = s[i-1].sum + num[i];
s[i].start = s[i-1].start;
s[i].end = i;
}
}
sort(s+1,s+n+1,cmp);
printf("Case %d:\n%d %d %d\n",k,s[1].sum,s[1].start,s[1].end);
if (k != T)
putchar('\n');
}
return 0;
}

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