Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 52833 Accepted Submission(s): 11726
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).
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.
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
scanf("%d",&T);
for(p=1;p<=T;p++) {
scanf("%d",&n);
max=-9999; //因为一个数a 是-1000~1000的,所以这里相当于变成最小值
t=0; //表示 某段连续和
sta=end=k=1; // sta最大和的开始,end最大和的结束,k记录每次求和的开始
for(i=1;i<=n;i++) {
scanf("%d",&a);
t+=a;
if(t>max) { //记录最大连续和的值
max=t;
sta=k;
end=i;
}
if(t<0) {
t=0;
k=i+1;
}
}
if(p!=1) printf("/n");
printf("Case %d:/n",p);
printf("%d %d %d/n",max,sta,end);
}
}
下面是用数组做的: