题目传送门
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
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
题意：
给定序列个数n及n个数，求该序列的最大连续子序列的和，要求输出最大连续子序列的和以及子序列的首位位置
分析：
常规求解子序列和问题，但是加了首尾位置sum维护当前首尾指向的区间和，maxi记录最大值 1、如果sum < 0，舍去即可，因为一定导致后面的值变小，不如直接选后面的值 2、如果sum > maxi，更新最大值注意结尾换行
AC代码：
#include <iostream>
#include <vector>
#include <utility>
#include <cstring>
#include <string>
#include <algorithm>
#include <map>
#include <queue>
#include <stack>
#include <cstdio>
#include <fstream>
#include <set>
using namespace std
;
typedef long long ll
;
const int INF
= 0x3f3f3f;
const int MAXN
= 1e5 + 100;
int a
[MAXN
];
int dp
[MAXN
];
int main() {
int T
;
cin
>> T
;
for (int z
= 1; z
<= T
; z
++) {
int n
;
cin
>> n
;
for (int i
= 0; i
< n
; i
++) {
cin
>> a
[i
];
}
int maxi
= -INF
;
int start
= 0;
int end
= 0;
int sum
= 0;
int ts
= 0, te
= 0;
while (end
< n
) {
sum
+= a
[end
];
if (sum
> maxi
) {
maxi
= sum
;
ts
= start
;
te
= end
;
}
if (sum
< 0) {
sum
= 0;
start
= end
+ 1;
}
end
++;
}
if (z
!= 1)
cout
<< endl
;
cout
<< "Case " << z
<< ":" << endl
<< maxi
<< " " << ts
+ 1 << " " << te
+ 1 << endl
;
}
return 0;
}
