Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 110957 Accepted Submission(s): 25601
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).
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.
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
Author
Ignatius.L
题意:求最大连续子序列和。
分析:
要找出和最大的子段,首先想到的是枚举,枚举的方法是,分别将数串中的每一个数作为子段的第一位数,然后子段长度依次递增。例如:
(2,-3,4,-1)这个数串枚举的所有情况为:
以2作为子段的第一位: 2,(2,-3),(2,-3,4),(2,-3,4,-1)。
以-3作为子段的第一位: -3,(-3,4),(-3,4,-1)。
以4作为子段的第一位: 4,(4,-1)。
以-1作为子段的第一位: -1 。
从枚举的过程中发现:
当前子段和的值为负值时,就没必要再往下进行,而应将下一位数作为子段的第一位。也就是说,当处理第i个数时,如果以第i-1个数为结尾的子段的和为正数,则不必将第i个数作为新子段的首位进行枚举,因为新子段加上前面子段和所得的正数,一定能得到更大的子段和。
代码:
#include<cstdio> int main() { int t,n,k,sum,temp_position,start,endd,num; scanf("%d",&t); for(k=1;k<=t;k++) { scanf("%d",&n); sum=0; temp_position=1; int ans=-1010; for(int i=1;i<=n;i++) { scanf("%d",&num); sum += num; if(sum > ans) { ans=sum; start=temp_position; endd=i; } if(sum<0) { sum=0; temp_position=i+1; } } printf("Case %d:\n",k); printf("%d %d %d\n",ans,start,endd); if(k!=t) printf("\n"); } return 0; }
感想:
1、这题其实真的很能反映“动态规划”最优子结构的特点。。。
2、动规的题,我感觉这只能算是第2题。。。还在找感觉。。。