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
程序分析：此题的基本思路是这样的。给你一串数，关键是选出其中和最大的连续的一串。可以这样想，首先选定第一个数，再与第二个数相比较，如果两都相加比第一个数小，就可以从第二个开始计算，如果相加结果变大，则还是从第一个相比较。之后再将每一个和与最大的和max相比较，大则交换，不大于则不变：
程序代码：
#include<stdio.h> int main() { int n,i,j,k,nums,a[100001],sum,b,start,end,start1,end1; scanf("%d",&n); for(i=1;i<=n;i++) { scanf("%d",&nums); for(j=1;j<=nums;j++) { scanf("%d",&a[j]); } sum=a[1]; b=-1001; start=end=0; for(k=1;k<=nums;k++) { if(b>=0) { b+=a[k]; end++; } else { b=a[k]; start=end=k; } if(b>=sum) { sum=b; start1=start; end1=end; } } printf("Case %d: ",i); printf("%d %d %d ",sum,start1,end1); if(i<n)printf(" "); } return 0; }