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:
这道题我 WA 3次 PE 1次 ==
7 1 6
思路：最大连续子序列：状态方程:sum[i]=max(sum[i-1]+a[i],a[i]);最后从头到尾扫一边
感言：这是一道一粗心就错的题目，假设你还找不到自己代码的错误。那么试一下下面測试数据：
4 0 0 5 0 5 1 3
4 0 0 -1 0 0 1 1
5 -7 -8 -9 -2 -3 -2 4 4
5 -1 -2 5 -2 8 11 3 5
AC代码：
#include<stdio.h>
int a[100005];
int main()
{
int maxx,sum,i,cnt,flag,tot,t,ms,me,T;
scanf("%d",&t);
cnt=1;
T=t;
while(t--)
{
int n;
scanf("%d",&n);
for(i=0;i<n;i++)
scanf("%d",&a[i]);
sum=0;
maxx=a[0];
flag=tot=ms=0;
me=1;
for(i=0;i<n;i++)
{
if(sum<0){
sum=a[i];
flag=i;
tot=i+1;
}
else{
tot++;
sum+=a[i];
}
if(sum>maxx){
ms=flag;
me=tot;
maxx=sum;
}
}
printf("Case %d:\n",cnt++);
printf("%d %d %d\n",maxx,ms+1,me);
if(cnt<=T)
printf("\n");
}
return 0;
}
这道题我 WA 3次 PE 1次 ==
本文转自mfrbuaa博客园博客，原文链接：http://www.cnblogs.com/mfrbuaa/p/5267815.html，如需转载请自行联系原作者
