Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 285219 Accepted Submission(s): 67716
Total Submission(s): 285219 Accepted Submission(s): 67716
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).
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.
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
本来想用常规方法做的,结果发现超时了。。。
for(int i=1;i<=n;i++){ int x; scanf("%d",&x); a[i]=a[i-1]+x; for(int j=0;j<i;j++){ if(max1<a[i]-a[j]){ max1=a[i]-a[j]; ue=i; ub=j; } } }
因为数据范围是1e5,所以用二重循环会超时,然后在网上看到用dp写的
感觉很巧妙
int k=1,s=0,e=0,summax=-1000,sum=0; for(i=0;i<n;i++) { scanf("%d",&a[i]); } for(i=0;i<n;i++) { sum+=a[i]; if(sum>summax) { summax=sum; s=k; e=i+1; } if(sum<0) { sum=0; k=i+2; } }
因为是连续的,所以用一个sum来记录连续的和
如果sum>0的话,与summax比较记录 开始,结束 位置
而如果sum<0的话,那么如果用这个sum与后面相加的话,肯定很亏,所以就吧sum赋值为0,把开始点变成下一位
嗯,很巧妙。。。。