杭电acm1003
http://acm.hdu.edu.cn/showproblem.php?pid=1003
Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 49213 Accepted Submission(s): 10950
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:
#include <iostream>
using namespace std;
int main()
{
int T,N,num,startP,endP;
cin>>T;
for(int k=0;k<T;k++)
{
cin>>N;
int max=-1001,sum=0,temp=1;
for(int i=0;i<N;i++)
{
cin>>num;
sum+=num;
if(sum>max)
{
max=sum;
startP=temp;
endP=i+1;
}
if(sum<0)
{
sum=0;
temp=i+2;
}
}
cout<<"Case "<<k+1<<":"<<endl<<max<<" "<<startP<<" "<<endP<<endl;
if(k!=T-1) cout<<endl;
}
return 0;
}
#include<stdio.h>
int main()
{
int T,N,i,j,startP,endP,max,num,sum,temp;
scanf("%d",&T);
for(j=0;j<T;j++)
{
max=-1001;
temp=1;sum=0;
scanf("%d",&N);
for(i=0;i<N;i++)
{
scanf("%d",&num);
sum+=num;
if(sum>max)
{
max=sum;
startP=temp;
endP=i+1;
}
if(sum<0)
{
sum=0;
temp=i+2;
}
}
printf("Case %d:\n%d %d %d\n",j+1,max,startP,endP);
if(j<T-1)
printf("\n");
}
return 0;
}
7 1 6