Maximum sum
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 30704 | Accepted: 9408 |
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.
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.
Source
POJ Contest,Author:Mathematica@ZSU
题意：
给一个数列，求出数列中不相交的两个字段和，要求和最大。
思路:
对每一个i来说，求出【0~i-1】的最大子段和以及【i~n-1】的最大子段和，再加起来求最大的一个就行了。[0~i-1]的最大子段和从左向右扫描，【i~n-1】从右想左扫描
即可，时间复杂度O(n).
代码：
1 #include<iostream> 2 #include<cstdio> 3 #define maxn 50001 4 #include<algorithm> 5 using namespace std; 6 int a[maxn]; 7 int left[maxn]; 8 int right[maxn]; 9 int max(int a,int b) 10 { 11 return a>b?a:b; 12 } 13 int main() 14 { 15 int t,i; 16 scanf("%d",&t); 17 while(t--) 18 { 19 int n; 20 scanf("%d",&n); 21 for(i=0;i<n;i++) 22 scanf("%d",&a[i]); 23 //此时::left【i】为包涵i最大字段和 24 ::left[0]=a[0]; 25 for( i=1; i<n;i++) 26 if(::left[i-1]<0) 27 ::left[i]=a[i]; 28 else 29 ::left[i]=::left[i-1]+a[i]; 30 //此时left[i]为i左边最大字段和 31 for(i=1; i<n;i++) 32 ::left[i]=max(::left[i],::left[i-1]); 33 ::right[n-1]=a[n-1]; 34 for(i=n-2;i>=0;i--) 35 { 36 if(::right[i+1]<0) 37 ::right[i]=a[i]; 38 else 39 ::right[i]=::right[i+1]+a[i]; 40 } 41 for(i=n-2;i>=0;i--) 42 ::right[i]=max(::right[i+1],::right[i]); 43 int res=-100000000; 44 for(i=1;i<n;i++) 45 { 46 res=max(res,::left[i-1]+::right[i]); 47 } 48 printf("%d ",res); 49 } 50 return 0; 51 }