HDU-1003 Max Sum(dp)
Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 285141 Accepted Submission(s): 67698
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
Author
Ignatius.L
Recommend
大体意思就是让你找到和最大的子串~~这样的话我们来用sum来表示前n项数的和是多少~~如果sum加完后小于0，那么就代表着，这前n项和对后续的影响是负担~做负功的感觉差不多~~所以我们直接舍去这前n项~~重下一项开始计就好了；相反的~加入这时sum>0就代表着前n项的加到后面对于到时候的前几项和是有利的（提升）继续保留就好。
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
int te;
cin>>te;
int Ca=1;
while(te--)
{
int n;
cin>>n;
int sum=0;
int maxn=-9999999;
int mst=1,men=1,st=1;
for(int s=1;s<=n;s++)
{
int a;
scanf("%d",&a);
sum+=a;
if(sum>maxn)
{
maxn=sum;
mst=st;
men=s;
}
if(sum<0)
{
sum=0;
st=s+1;
}
}
printf("Case %d:\n",Ca++);
printf("%d %d %d\n",maxn,mst,men);
if(te>=1)
{
cout<<"\n";
}
}
return 0;
}
