POJ-2593 Max Sequence(不连续的最大子段和)
Max Sequence
Time Limit: 3000MS | Memory Limit: 65536K | |
Total Submissions: 16001 | Accepted: 6715 |
Description
Give you N integers a1, a2 ... aN (|ai| <=1000, 1 <= i <= N).
![](http://poj.org/images/2593_1.jpg)
You should output S.
![](http://poj.org/images/2593_1.jpg)
You should output S.
Input
The input will consist of several test cases. For each test case, one integer N (2 <= N <= 100000) is given in the first line. Second line contains N integers. The input is terminated by a single line with N = 0.
Output
For each test of the input, print a line containing S.
Sample Input
5 -5 9 -5 11 20 0
Sample Output
40
大致题意:给出一个数列,求出数列中不相交的两个子段和,要求子段和最大。
1 #include <stdio.h> 2 int main() 3 { 4 int left[100005]; 5 int right[100005]; 6 int a[100005]; 7 while(1) 8 { 9 int n,i; 10 scanf("%d",&n); 11 if(n==0) 12 break; 13 for(i=0;i<n;i++) 14 scanf("%d",&a[i]); 15 int sumL=0; 16 int max=-99999999; 17 for(i=0;i<n;i++)//从左向右求到i结尾的最大连续子串的最大值 //如果这个循环看不懂的话,可以先做下hdu1003 18 { 19 sumL=sumL+a[i]; 20 if(sumL>max) 21 max=sumL; 22 if(sumL<0) 23 sumL=0; 24 left[i]=max; 25 } 26 max=-99999999; 27 int tmp=-99999999; 28 int sumR=0; 29 for(i=n-1;i>=1;i--) 30 { 31 sumR=sumR+a[i]; 32 if(sumR>max) 33 max=sumR; 34 if(sumR<0) 35 sumR=0; 36 if(tmp<max+left[i-1]) 37 tmp=max+left[i-1]; 38 } 39 printf("%d\n",tmp); 40 } 41 return 0; 42 }