标签：
Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 208486 Accepted Submission(s): 48797
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
细心+认真+恒心
#include<stdio.h>
#include<string.h>
#define max(a,b) a>b?a:b
int a[100005],d[100005];
int main(){
int t,n,ant=1;
int i,j1,j2,maxx;
scanf("%d",&t);
while(t--){
memset(d,0,sizeof(d));
scanf("%d",&n);
for( i=1;i<=n;i++)
scanf("%d",&a[i]);
maxx=d[1]=a[1]; //初始值
j1=j2=1; //初始位置
for( i=2;i<=n;i++){
d[i]=max(a[i],d[i-1]+a[i]); //当前值与和比较，较大的放入d中
if(maxx<d[i]) {maxx=d[i];j2=i;} //找出最大的和，并记录位置
}
for( i=2;i<=j2;i++)
if(d[i-1]<0) j1=i;//找出最大和位置前，最后一次改变开始值的位置
if(t==0) printf("Case %d:\n%d %d %d\n",ant++,maxx,j1,j2);
else printf("Case %d:\n%d %d %d\n\n",ant++,maxx,j1,j2);
}
return 0;
}
hdu 1003 Max Sum
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原文地址：http://blog.csdn.net/z_huing/article/details/51354950
