HDU 1003 MAXSUM(最大子序列和)
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
题目主要是求连续最大子序列的和,并输出最大子序列的左边界和右边界。
注意输出的格式为每两组数据之间输出一个空行,最后一组数据没有。
1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 #include <algorithm> 5 using namespace std; 6 #define INF 0x3f3f3f3f 7 int main() 8 { 9 int t,n; 10 int i,j,k=0; 11 int Max,sum,x,y,l,d; 12 scanf("%d",&t); 13 while(t--) 14 { 15 Max=sum=-INF; 16 scanf("%d",&n); 17 for(i=1; i<=n; i++) 18 { 19 scanf("%d",&d); 20 if(sum+d<d) 21 sum=d,l=i; 22 else 23 sum+=d; 24 if(Max<sum) 25 { 26 x=l; 27 y=i; 28 Max=sum; 29 } 30 } 31 if(k) 32 printf("\n"); 33 printf("Case %d:\n",++k); 34 printf("%d %d %d\n",Max,x,y); 35 } 36 return 0; 37 }