Program3_A

我现在做的是第三专题编号为1001的试题， 具体内容如下所示：

Problem A

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 182 Accepted Submission(s) : 30
[align=left]Problem Description[/align]
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.<br>

[align=left]Input[/align]
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).<br>

[align=left]Output[/align]
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.<br>

[align=left]Sample Input[/align]

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

[align=left]Sample Output[/align]

Case 1:
14 1 4

Case 2:
7 1 6

简单题意：

在组成环的n个数中求出不超过m个连续的数相加之和最大时的最大值、起始位置和终止位置（如果大于n则要减去n）

解题思路：

设一数组sum，sum[i]含义：所有以a[i]为结尾的序列的序列和构成一个集合，此集合的最大值就是sum[i]

如1,2,3.所有以a[2]=3为结尾的序列的序列和集合是{6,5,3},因而sum[2]=6.
sum的状态转移方程：sum[i] = max{sum[i-1]+a[i], a[i]}

ans必定是sum[0···(k-1)]之一。由于要记录起始位置和结束位置，引入s数组记录获得sum的序列的起始元素的位置，而由sum的定义，sum[i]的结束位置是i不用另外记录。

#include <cstdio>

using namespace std;

int a[100005], sum[100005], s[100005];

int main()

{

int T, N;

int ca = 0, ans;

scanf("%d", &T);

while(ca < T)

{

scanf("%d", &N);

for(int i = 0; i < N; i++)

scanf("%d", &a[i]);

ans = 0;

sum[0] = a[0];

s[0] = 0;

for(int i = 1; i < N; i++)

{

if(sum[i-1] >= 0)

{

sum[i] = sum[i-1] + a[i];

s[i] = s[i-1];

}

else

{

sum[i] = a[i];

s[i] = i;

}

if(sum[ans] < sum[i])

ans = i;

}

printf("Case %d:\n%d %d %d\n", ++ca, sum[ans], s[ans] + 1, ans + 1);

if(ca != T) puts("");

}

return 0;

}