Max Sum(最大连续子序列)

Max Sum

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status

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

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

Sample Output

Case 1:14 1 4Case 2:7 1 6

#include<iostream>#include<iostream>#include<string.h>#include<algorithm>using namespace std;int a[1000000],n,t,d[1000000];int main(){int k=1;scanf("%d",&t);while(t--){scanf("%d",&n);for(int i=1;i<=n;i++)scanf("%d",&a[i]);d[1]=a[1];for(int i=2;i<=n;i++){if(d[i-1]<0)d[i]=a[i];elsed[i]=a[i]+d[i-1];} int max1=d[1]; int end=1; for(int i=2;i<=n;i++) {if(max1<d[i]){max1=d[i];end=i;} } int r=0,f=end; for(int i=end;i>=1;i--) { r+=a[i]; if(r==max1) f=i; } printf("Case %d:\n%d %d %d\n",k++,max1,f,end); if(t!=0) puts("");}}