poj 2479 Maximum sum
Maximum sum
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 24408 | Accepted: 7404 |
Description
Given a set of n integers: A={a1, a2,..., an}, we define a function d(A) as below:
Your task is to calculate d(A).
Input
The input consists of T(<=30) test cases. The number of test cases (T) is given in the first line of the input.
Each test case contains two lines. The first line is an integer n(2<=n<=50000). The second line contains n integers: a1, a2, ..., an. (|ai| <= 10000).There is an empty line after each case.
Each test case contains two lines. The first line is an integer n(2<=n<=50000). The second line contains n integers: a1, a2, ..., an. (|ai| <= 10000).There is an empty line after each case.
Output
Print exactly one line for each test case. The line should contain the integer d(A).
Sample Input
1 10 1 -1 2 2 3 -3 4 -4 5 -5
Sample Output
13
Hint
In the sample, we choose {2,2,3,-3,4} and {5}, then we can get the answer.
Huge input,scanf is recommended.
Huge input,scanf is recommended.
#include <iostream>
using namespace std;
int array[100001], num[100001];
const int MIN = INT_MIN;
int main()
{
int n;
int tmp, ans, i, sum;
int cases;
cin>>cases;
while(cases--)
{
cin>>n;
tmp = MIN; sum = 0;
for(i = 1; i <= n; i++){
scanf("%d", &num[i]);
sum += num[i];
if(sum > tmp)
tmp = sum;
array[i] = tmp;
if(sum < 0)
sum = 0;
}
tmp = ans = MIN;
sum = 0;
for(i = n; i > 1; i--){
sum += num[i];
if(sum > tmp)
tmp = sum;
if(ans < (array[i-1]+tmp))
ans = array[i-1]+tmp;
if(sum < 0)
sum = 0;
}
cout<<ans<<endl;
}
return 0;
}
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