Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 206582 Accepted Submission(s):
48294
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.
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).
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.
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
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5
Sample Output
Case 1:
14 1 4
14 1 4
Case 2:
7 1 6
Author
Ignatius.L
Recommend
题意：求连续子序列和的最大值。
题解： 注意这道题要求是输出第一个符合条件的序列，而且存在负数，样例之间要用空行分隔。
#include <bits/stdc++.h>
using namespace std;
const int maxn=1e5+;
int a[maxn],n;
int main()
{
int t,cas=;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
int sum=,ans=-;
int s=,e=,k=;
for(int i=;i<=n;i++)
{
scanf("%d",&a[i]);
sum+=a[i];
if(sum>ans)
{
s=k;
e=i;
ans=sum;
}
if(sum<) //0的意义就是这段数做的是负功
{
sum=;
k=i+;
}
}
printf("Case %d:\n",cas++);
printf("%d %d %d\n",ans,s,e);
if(t>) puts("");
}
return ;
}
#include <bits/stdc++.h>
using namespace std;
const int N=1e5+;
int a[N],n;
int main()
{
int t,cas=;
cin>>t;
while(t--)
{
cin>>n;
int sum=,mx=-,s=,e=,ts=;
for(int i=;i<=n;i++)
{
cin>>a[i];
sum+=a[i];
if(sum>mx)
{
mx=sum;
s=ts;
e=i;
}
if(sum<)
{
sum=;
ts=i+;
}
}
printf("Case %d:\n",cas++);
printf("%d %d %d\n",mx,s,e);
if(t) printf("\n");
}
return ;
}
/*
100
2 1 2
1 1
3 -1 1 2
2 -7 3
*/
牢记顺序是 加 大于 小于！！