Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 67750 Accepted Submission(s): 15546
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
这是一个最最大子段和的问题。并且输出子段的头根尾部。
#include<iostream>
#include<stdio.h>
using namespace std;
int a[100000+10];
int sum[100000+10];
void maxsum(int n)
{
sum[0]=-1;
sum[1]=a[1];
int max=a[1];
int right=1;
for(int i=2;i<=n;i++)
{
if(sum[i-1]>0)sum[i]=sum[i-1]+a[i];
else sum[i]=a[i];
if(max<sum[i])
{
max=sum[i];right=i;
}
}
int left=right;
while(sum[left]>=0)left--;
if(left!=right)//注意这里要不然rihgt就会大于left了。
left++;
printf("%d %d %d\n",max,left,right);
}
int main()
{
int t;
scanf("%d",&t);
for(int i=1;i<=t;i++)
{
int n;
scanf("%d",&n);
for(int j=1;j<=n;j++)
{
scanf("%d",&a[j]);
}
printf("Case %d:\n",i);
maxsum(n);
if(i!=t)printf("\n");
}
return 0;
}
371

被折叠的 条评论
为什么被折叠?



