1、http://poj.org/problem?id=2479
2、题目大意:
给定一串数字,求连续的两个不相交的子段的数和最大,由题目不难看出此题求得是两个子段的数和
3、思路:跟上一篇博客hdu 1024 一样的想法,只是稍微处理一下题目,就不难看出类型一样
4、题目:
Maximum sum
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 28823 | Accepted: 8809 |
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
4、代码:
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define N 50005
int a[N];
int now[N];
int pre[N];
int main()
{
int t,n,maxx;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
memset(now,0,sizeof(now));
memset(pre,0,sizeof(pre));
for(int i=1;i<=2;i++)
{
maxx=-1*N*10000;
for(int j=i;j<=n;j++)
{
now[j]=max(now[j-1]+a[j],pre[j-1]+a[j]);
pre[j-1]=maxx;
if(now[j]>maxx)
maxx=now[j];
}
}
printf("%d\n",maxx);
}
return 0;
}
/*
1
10
1 -1 2 2 3 -3 4 -4 5 -5
*/