【思维】poj1003 Max Sum

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 242921    Accepted Submission(s): 57358


Problem Description

Given a sequence a[1],a[2],a[3]......a
, 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

 

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题意：找一个连续的最长子序列使得区间和最大；

思路：这道题在2017河工大校赛中出现过，相比本题不需要输出区间的起始位置和终止位置；当时没有A出，解题思路都是一样的；正着找一遍，反者着一边；WA了好几次是因为初始化的问题；

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=100005;
const int inf=0x3f3f3f3f;

struct p
{
int s,t,c;
}e[maxn];
int a[maxn];

bool cmp(p e1,p e2)
{
if(e1.c!=e2.c)
return e1.c>e2.c;
else
return e1.s<e2.s;
}

int main()
{
int T,t=0;
scanf("%d",&T);
while(T--)
{
t++;
int n;
scanf("%d",&n);
for(int i=0;i<n;i++)
scanf("%d",&a[i]);

memset(e,-0x3f,sizeof(e));
int sum=0,k=0;
e[k].s=0;
int maxx=-inf;
for(int i=0;i<n;i++)
{
sum+=a[i];
if(sum>=maxx)
{
maxx=sum;
e[k].t=i;
e[k].c=sum;
}
else if(sum<0)
{
sum=0;
e[++k].s=i+1;
}
}

k++;
e[k].t=n-1,sum=0;
for(int i=n-1;i>=0;i--)
{
sum+=a[i];
if(sum>=maxx)
{
maxx=sum;
e[k].s=i;
e[k].c=sum;
}
else if(sum<0)
{
sum=0;
e[++k].t=i-1;
}
}
sort(e,e+k+1,cmp);

//for(int i=0;i<=k;i++)
//printf("%d %d %d\n",e[i].s,e[i].t,e[i].c);

printf("Case %d:\n%d %d %d\n",t,e[0].c,e[0].s+1,e[0].t+1);
if(T>=1)  printf("\n");
}
return 0;
}